Math 7 · 4.MD.C.7

Congruent angles, angle addition and bisectors

Two angles sit in different places, a ray splits an angle into pieces, a fold halves a corner. When are two angles the same size, how do the pieces add up to the whole, and what does a ray that cuts an angle exactly in half tell you?

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Measuring anglesGrades 4 to 8

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Why this exists

Look at the two angles in Figure 1. One has short arms and opens up and to the right. The other has arms nearly twice as long and opens to the left. Are they the same?

They are not the same angle: they sit in different places, and one is made of different rays from the other. But they open by exactly the same amount, and in almost every problem that is the only thing that matters. Geometry needs a word for “same size and shape, wherever it is drawn”, so that it can say this without pretending the two angles are one object. That word is congruent, and angles are the first place most people meet it. The same word does the same job for segments (same length), and later for whole triangles, where it carries most of the weight of a proof.

Two more facts about angle sizes are needed before any real angle problem can be written down as a proof. You already know that angle measures add: put two angles side by side and the angle around both is the total. But “it looks like they add” is not a reason anyone can write down, and every step of a proof has to point at something already agreed. The angle addition postulate is that agreed thing. It says exactly when two measures add to a third, and it is the one fact you cite every time you split an angle, glue two together, or work out a piece from the whole.

The third fact is about splitting an angle equally. Fold a paper corner so that one edge lies along the other, and the crease splits the corner into two equal parts. A carpenter joining two boards at a corner cuts each board along that same crease, so the two cut ends meet flush. The crease is the angle bisector, and it turns up far more often than the paper corner suggests, because it is the only line that treats the two sides of an angle exactly alike.

Definition

Congruent angles

Two angles are congruent when they have the same measure.

In symbols, \angle ABC \cong \angle DEF means exactly the same thing as

m\angle ABC = m\angle DEF

The two statements are about different kinds of thing. The left one compares two angles, which are figures made of rays; the right one compares two numbers, their measures in degrees. So \cong goes between angles and = goes between measures. Writing \angle ABC = \angle DEF would claim the two are the very same angle, made of the same two rays.

Figure 1Matching arcs mark the two angles congruent, however long their arms and whichever way they face.

On a figure, congruent angles are marked with matching arcs: the same number of arcs in each. In Figure 1 each angle carries one arc, which says \angle ABC \cong \angle DEF. If a figure also had a second pair of congruent angles, they would get two arcs each, so the pairs cannot be mixed up. Marks are information. A figure is not drawn to scale unless it says so, and two angles that merely look equal are not given as congruent until they carry matching marks.

The angle addition postulate

Take an angle \angle AOC with vertex O, smaller than a straight angle. Its interior is the region between its two sides: the points that are on the same side of line OA as C, and on the same side of line OC as A. In Figure 2 it is the shaded wedge, carried on outward for ever.

Figure 2Ray \overrightarrow{OB} runs inside \angle AOC and splits it into two pieces.

Angle Addition Postulate. If a point B lies in the interior of \angle AOC, then

m\angle AOB + m\angle BOC = m\angle AOC

In words: when a ray from the vertex runs inside an angle, the two smaller angles it makes add up to the whole angle. The two smaller angles are adjacent: they share the vertex O and the side \overrightarrow{OB}, and nothing else.

It is called a postulate because it is not proved from anything simpler. It is one of the starting rules that school geometry agrees on, alongside the rule that every angle has a measure between 0^\circ and 180^\circ. It is the angle version of the segment addition postulate: there a point between two others splits a length, here a ray inside an angle splits a measure.

Angle bisector

A ray \overrightarrow{BD} bisects \angle ABC when D lies in the interior of the angle and the two angles it makes are congruent:

\angle ABD \cong \angle DBC

The ray is then called the bisector of \angle ABC. In Figure 3 the matching ticks mark the two equal halves.

Figure 3\overrightarrow{BD} bisects \angle ABC. Matching ticks mark the two equal halves.

Two things in the definition do real work. The ray must start at the vertex B, since a ray starting anywhere else makes no angle with the sides at B. And D must be inside the angle, between the two sides, so that the two halves really are parts of the whole. Boundaries below shows what goes wrong without that second condition.

A line or a segment through the vertex that contains the bisecting ray is also said to bisect the angle. In a triangle, for instance, the bisector of an angle is usually drawn as a segment that stops at the opposite side.

Intuition

An angle is an amount of opening, the turn from one arm to the other. Three things about how it is drawn make no difference to that amount.

Where it is. Sliding an angle across the page does not open or close it.
How long the arms are. An arm is a ray, which goes on for ever; a drawing just stops it somewhere. Longer arms do not make a bigger angle.
Which way it faces. Turning an angle, or looking at it in a mirror, leaves the opening unchanged.

So congruent angles are exactly the angles you could make fit: trace one on thin paper, slide it and turn it (flip the paper over if you like), and it lies exactly on top of the other. That is the old way of comparing angles without measuring, and Why it is true shows it agrees with the definition by measure.

Thinking of an angle as a turn also explains the addition postulate. To get from \overrightarrow{OA} to \overrightarrow{OC} you can turn in one go, or you can stop at \overrightarrow{OB} on the way and then carry on. The total turn is the same either way, so the first leg plus the second leg equals the whole. The condition that B is inside is what makes \overrightarrow{OB} a stop on the way. If \overrightarrow{OB} pointed off somewhere outside the angle, getting there and back would not be a route from \overrightarrow{OA} to \overrightarrow{OC} at all, and there would be no reason for the measures to add.

The bisector is the stop exactly halfway. Stand at the vertex and face along it: one side of the angle is off to your left, the other off to your right, and both lean away from you by exactly the same amount. That is the fold in Figure 4. Fold the page along the bisector and each side of the angle lands exactly on the other. A point P on one side, a given distance from B, lands on the point Q on the other side at that same distance. The crease cuts \overline{PQ} in half and meets it at a right angle, which is what a fold always does to a point and its image.

Figure 4Fold along the bisector and each side lands on the other. P lands on Q.

So the bisector is the angle’s line of symmetry, and a reflection across it swaps the two sides. Every point on it sits the same distance from both sides, which is why it is the tool for finding the center of a circle that touches two walls at once.

Results

Congruence

Congruence of angles behaves like equality. For any angles \angle A, \angle B and \angle C:

Reflexive. \angle A \cong \angle A.
Symmetric. If \angle A \cong \angle B, then \angle B \cong \angle A.
Transitive. If \angle A \cong \angle B and \angle B \cong \angle C, then \angle A \cong \angle C.

Each is inherited from the matching property of = between measures. They are what let a proof cite a shared angle or a chain of congruences.

Same measure means one fits on the other. Two angles are congruent exactly when some sequence of slides and turns carries one onto the other. No flip is ever needed.

All right angles are congruent. Every right angle measures 90^\circ, so any two of them are congruent. See Proving angle pair theorems.

Supplements of congruent angles are congruent. If \angle 1 \cong \angle 2, and \angle 3 is supplementary to \angle 1 while \angle 4 is supplementary to \angle 2, then

\angle 3 \cong \angle 4

The same holds for complements of congruent angles (Exercise 42).

Angle addition

For B in the interior of \angle AOC, the postulate can be read three ways, depending on which measure you are missing:

m\angle AOC = m\angle AOB + m\angle BOC

m\angle AOB = m\angle AOC - m\angle BOC

m\angle BOC = m\angle AOC - m\angle AOB

Each piece is smaller than the whole. Both pieces have positive measure, so m\angle AOB < m\angle AOC.

It works for any number of pieces. If rays \overrightarrow{OB} and \overrightarrow{OC} both lie inside \angle AOD, in the order OA, OB, OC, OD, then applying the postulate twice gives

\begin{aligned} m\angle AOD &= m\angle AOB + m\angle BOC \\ &\quad + m\angle COD \end{aligned}

Bisectors

Half the angle. If \overrightarrow{BD} bisects \angle ABC, then each half measures half of the whole:

m\angle ABD = m\angle DBC = \tfrac{1}{2}\, m\angle ABC

One bisector. Every angle smaller than a straight angle has exactly one bisector.

Bisectors of a linear pair. If two angles form a linear pair, their bisectors are perpendicular.

Why it is true

Same measure means one fits on the other. Take two angles with the same measure, \angle ABC and \angle DEF.

First slide the whole of \angle ABC so that B lands on E. A slide moves every point the same distance in the same direction, so it does not change the opening. Now the two angles share a vertex, as in Figure 5.

Figure 5Slide B onto E, then turn. Because the openings match, the second ray lands too.

Next turn about E until ray \overrightarrow{BA} lies along \overrightarrow{ED}. A turn about the vertex moves both rays by the same amount, so the opening is still the same. Ray \overrightarrow{BC} now makes an angle equal to m\angle DEF with \overrightarrow{ED}, so it points one of only two ways: along \overrightarrow{EF}, or into the mirror position on the other side of \overrightarrow{ED}.

If it points along \overrightarrow{EF}, the angles coincide. If it lands in the mirror position, undo the turn and turn the other ray onto \overrightarrow{ED} instead: the angle’s two arms swap roles, and now the first arm lands on \overrightarrow{EF}. An angle is just its two rays, with neither one first, so either way the whole angle lies on the whole angle. That is why no flip is needed.

The converse is quicker. A slide or a turn never changes a measure, so if one angle can be carried onto another, their measures were equal all along.

Supplements of congruent angles. Suppose \angle 1 \cong \angle 2, so m\angle 1 = m\angle 2. Supplementary angles add to 180^\circ, so

\begin{aligned} m\angle 3 &= 180^\circ - m\angle 1 \\ m\angle 4 &= 180^\circ - m\angle 2 \end{aligned}

The right-hand sides are equal, because they subtract equal numbers from 180^\circ. So m\angle 3 = m\angle 4, which is \angle 3 \cong \angle 4. The proof for complements is the same with 90^\circ in place of 180^\circ.

The angle addition postulate. A postulate is accepted rather than proved, but it is worth seeing why this one is a sensible thing to accept, because the reason is the protractor. Lay a protractor with its center on O and its zero line along \overrightarrow{OA}. Every ray from O on that side of line OA now crosses the scale at one reading, and the measure of the angle between two rays is the difference of their readings. Say \overrightarrow{OC} reads c. Because B is inside \angle AOC, \overrightarrow{OB} crosses the scale somewhere between 0 and c; call that reading b. Then

m\angle AOB = b - 0 = b \qquad\qquad m\angle BOC = c - b

and adding them,

b + (c - b) = c = m\angle AOC

So the postulate is what you get by agreeing that measuring an angle means reading a scale. School geometry packages exactly that agreement as the protractor postulate, and the angle addition postulate comes with it.

Some treatments go further and make the rule hold without any condition at all. In Anton Petrunin’s axioms for the plane, an angle’s measure carries a sign (turning one way counts as positive, the other way as negative), and for any three rays from O the measures of \angle AOB and \angle BOC add to the measure of \angle AOC up to a whole number of full turns. The price is signed measures and arithmetic that wraps round at 360^\circ. The school version pays instead with the word “inside”, and keeps every measure an ordinary number between 0^\circ and 180^\circ.

Half the angle. Because D lies inside \angle ABC, the angle addition postulate says the two parts add to the whole:

m\angle ABD + m\angle DBC = m\angle ABC

The two parts are congruent, so they have the same measure. Replace m\angle DBC with m\angle ABD:

\begin{aligned} 2\, m\angle ABD &= m\angle ABC \\ m\angle ABD &= \tfrac{1}{2}\, m\angle ABC \end{aligned}

One bisector. On one side of the line through B and A, there is exactly one ray from B that makes any chosen angle with \overrightarrow{BA}; this is the fact that makes a protractor work. A bisector must lie on the same side as C, because D is inside the angle, and it must make an angle of exactly \tfrac{1}{2}\, m\angle ABC with \overrightarrow{BA}. Only one ray does both.

Bisectors of a linear pair. Let the linear pair measure x^\circ and (180 - x)^\circ. Each bisector takes half of its own angle, and the two halves that sit between the bisectors are next to each other, so the angle between the bisectors is

\frac{x}{2} + \frac{180 - x}{2} = \frac{180}{2} = 90

degrees. The value of x cancels completely, which is why the result holds for every linear pair and not only for the ones drawn in a book.

Worked examples

Example 1

Sorting angles by measure

Which of the five angles in Figure 6 are congruent to each other?

Figure 6Five angles to sort. Only the measure decides.

Solution

Only the measures count. Ignore the arm lengths and the directions the angles face.

\angle P, \angle R and \angle T all measure 40^\circ, so

\angle P \cong \angle R \cong \angle T

\angle R has much longer arms than \angle P, and \angle T opens the other way, but neither fact changes a measure.

\angle Q measures 65^\circ and \angle S measures 140^\circ. Neither matches any other angle, so neither is congruent to any of the others.

Notice \angle S in particular. It is related to \angle P, since 180^\circ - 140^\circ = 40^\circ makes the two supplementary. Supplementary is a different relationship from congruent: supplementary angles add to 180^\circ, congruent angles are equal.

Example 2

Adding the pieces, and finding a missing one

(a) In Figure 7, point B lies in the interior of \angle AOC, with m\angle AOB = 35^\circ and m\angle BOC = 50^\circ. Find m\angle AOC.

Figure 7Two pieces are given. Find the whole.

(b) Point S lies in the interior of \angle PQR. If m\angle PQR = 128^\circ and m\angle PQS = 47^\circ, find m\angle SQR.

Solution

(a) B is in the interior of \angle AOC, so the angle addition postulate applies to these three angles:

\begin{aligned} m\angle AOC &= m\angle AOB + m\angle BOC \\ &= 35^\circ + 50^\circ = 85^\circ \end{aligned}

85^\circ is larger than each piece, as a whole must be.

(b) Because S is inside \angle PQR, the postulate says the two pieces add to the whole:

m\angle PQS + m\angle SQR = m\angle PQR

Here the whole is known and one piece is missing. Put in the two known measures and subtract 47^\circ from both sides:

\begin{aligned} 47^\circ + m\angle SQR &= 128^\circ \\ m\angle SQR &= 81^\circ \end{aligned}

Check. 47 + 81 = 128.

Example 3

Pieces written as expressions

In Figure 8, S lies in the interior of \angle PQR and m\angle PQR = 95^\circ. Find x and the measure of each piece.

Figure 8The two pieces written as expressions in x.

Solution

The postulate turns the picture into an equation: the two pieces add to the whole.

\begin{aligned} (2x + 5) + (3x - 10) &= 95 \\ 5x - 5 &= 95 \\ 5x &= 100 \\ x &= 20 \end{aligned}

Now substitute back. m\angle PQS = 2(20) + 5 = 45^\circ and m\angle SQR = 3(20) - 10 = 50^\circ.

Check. 45 + 50 = 95, and both pieces are positive, as angle measures must be.

Example 4

The whole as an expression too

Point M lies in the interior of \angle JKL. The measures are m\angle JKM = (3x + 6)^\circ, m\angle MKL = (2x + 13)^\circ and m\angle JKL = (8x - 14)^\circ. Find x and all three measures.

Solution

The whole being unknown as well changes nothing. The postulate still says piece plus piece equals whole, and that is one equation in x:

\begin{aligned} (3x + 6) + (2x + 13) &= 8x - 14 \\ 5x + 19 &= 8x - 14 \\ 33 &= 3x \\ x &= 11 \end{aligned}

Substitute x = 11 back into each expression:

\begin{aligned} m\angle JKM &= 3(11) + 6 = 39^\circ \\ m\angle MKL &= 2(11) + 13 = 35^\circ \\ m\angle JKL &= 8(11) - 14 = 74^\circ \end{aligned}

Check. 39 + 35 = 74. The whole comes out equal to the sum of the pieces, which it had to.

Example 5

Halving and doubling

\overrightarrow{BD} bisects \angle ABC and m\angle ABC = 84^\circ. Find m\angle ABD and m\angle DBC. Then, for a different angle \angle PQR bisected by \overrightarrow{QS}, you are told m\angle PQS = 29^\circ. Find m\angle PQR.

Solution

A bisector makes each half equal to half the whole:

m\angle ABD = m\angle DBC = \tfrac{1}{2}(84^\circ) = 42^\circ

For the second angle the question runs the other way. \angle PQS is one half, and the whole is two equal halves:

m\angle PQR = 2(29^\circ) = 58^\circ

Check. 42 + 42 = 84 and 29 + 29 = 58, as the angle addition postulate requires.

Example 6

Two halves as expressions

In Figure 9, \overrightarrow{BD} bisects \angle ABC, with m\angle ABD = (3x + 5)^\circ and m\angle DBC = (5x - 17)^\circ. Find x and m\angle ABC.

Figure 9The two halves of a bisected angle, written as expressions in x.

Solution

The two halves of a bisected angle are congruent, so by the definition of congruence their measures are equal. Set the expressions equal and solve:

\begin{aligned} 3x + 5 &= 5x - 17 \\ 22 &= 2x \\ x &= 11 \end{aligned}

Substitute into both expressions, since they must agree and checking both catches a slip:

3(11) + 5 = 38 \qquad\qquad 5(11) - 17 = 38

Each half is 38^\circ, and by angle addition the whole angle is the sum of its halves: m\angle ABC = 38^\circ + 38^\circ = 76^\circ.

The same step, setting two measures equal, settles any pair of congruent angles, whether or not they share a vertex.

Example 7

A half and the whole

In Figure 10, \overrightarrow{BD} bisects \angle ABC. Given m\angle ABD = (2x + 5)^\circ and m\angle ABC = (5x - 8)^\circ, find x, m\angle ABD and m\angle ABC.

Figure 10\overrightarrow{BD} splits \angle ABC into two congruent angles.

Solution

This time one expression is a half and the other is the whole, so they are not equal. Two facts are in play. The halves are congruent, so \angle DBC also measures (2x + 5)^\circ. And D is inside the angle, so by the angle addition postulate the two halves add to the whole:

\begin{aligned} (2x + 5) + (2x + 5) &= 5x - 8 \\ 4x + 10 &= 5x - 8 \\ x &= 18 \end{aligned}

Then m\angle ABD = 2(18) + 5 = 41^\circ and m\angle ABC = 5(18) - 8 = 82^\circ.

Check. 41 + 41 = 82, so the two congruent halves do fill the whole.

Reading which angle each expression names is the whole difficulty. Two halves are equal; a half and the whole are in the ratio 1 : 2.

Example 8

Three angles in a chain

\angle A \cong \angle B and \angle B \cong \angle C, with m\angle A = (x^2 - 3)^\circ, m\angle B = (3x + 7)^\circ and m\angle C = (4x + 2)^\circ. Find x and the common measure.

Solution

Start with the pair that gives a linear equation. From \angle B \cong \angle C:

3x + 7 = 4x + 2 \qquad\Longrightarrow\qquad x = 5

Now use \angle A \cong \angle B to test it: 5^2 - 3 = 22 and 3(5) + 7 = 22. They agree, so x = 5 and all three angles measure 22^\circ.

It is worth seeing why the first pair alone would not settle it. \angle A \cong \angle B gives x^2 - 3 = 3x + 7, so x^2 - 3x - 10 = 0 and (x - 5)(x + 2) = 0. The root x = -2 makes m\angle A = 1^\circ and m\angle B = 1^\circ, which is a genuine angle, but it makes m\angle C = 4(-2) + 2 = -6, and no angle has a negative measure. By transitivity \angle A \cong \angle C too, and only x = 5 satisfies all three at once.

Example 9

Supplements of a congruent pair

In Figure 11, \angle 1 \cong \angle 2. Also m\angle 3 = (2y + 16)^\circ and m\angle 4 = (3y - 9)^\circ. Find y and all four angles.

Figure 11\angle 1 \cong \angle 2 is given. The angles beside them follow.

Solution

\angle 1 and \angle 3 sit side by side along the line, so they are supplementary. So are \angle 2 and \angle 4. Supplements of congruent angles are congruent, so \angle 3 \cong \angle 4 and their measures are equal:

\begin{aligned} 2y + 16 &= 3y - 9 \\ y &= 25 \end{aligned}

Then m\angle 3 = 2(25) + 16 = 66^\circ and m\angle 4 = 3(25) - 9 = 66^\circ, which agree. Each of \angle 1 and \angle 2 is the supplement of a 66^\circ angle:

m\angle 1 = m\angle 2 = 180^\circ - 66^\circ = 114^\circ

The figure never gave a measure for \angle 1 or \angle 2. Knowing only that they are congruent was enough. Nothing in the argument needed the angles to be near each other, which is what makes it useful in a proof: two angles on opposite sides of a figure can be shown congruent because their supplements are.

Example 10

Overlapping angles

In Figure 12, four rays leave O in the order \overrightarrow{OA}, \overrightarrow{OB}, \overrightarrow{OC}, \overrightarrow{OD}, each of the middle two inside the angle formed by its neighbors. Show that if m\angle AOC = m\angle BOD, then m\angle AOB = m\angle COD.

Figure 12\angle AOC and \angle BOD overlap in the shaded piece \angle BOC.

Solution

The two big angles overlap: both contain the shaded piece \angle BOC. Split each one at the ray it contains.

B is in the interior of \angle AOC, so by the angle addition postulate

m\angle AOC = m\angle AOB + m\angle BOC

C is in the interior of \angle BOD, so by the postulate again

m\angle BOD = m\angle BOC + m\angle COD

The left sides are equal, because that is what we were given. So the right sides are equal:

m\angle AOB + m\angle BOC = m\angle BOC + m\angle COD

Subtract m\angle BOC from both sides:

m\angle AOB = m\angle COD

Take away the shared piece from two equal wholes and what is left over must match. This is the same move that proves vertical angles equal, where the shared piece is a neighboring angle and the two wholes are straight angles. For numbers: if m\angle AOC = m\angle BOD = 80^\circ and m\angle BOC = 48^\circ, the outer pieces are both 80^\circ - 48^\circ = 32^\circ, which is how Figure 12 is drawn.

Example 11

Bisecting twice

m\angle ABC = 112^\circ. \overrightarrow{BD} bisects \angle ABC, and \overrightarrow{BE} bisects \angle DBC. Find m\angle ABE and m\angle EBC.

Solution

Work from the outside in. The first bisector halves the whole angle:

m\angle ABD = m\angle DBC = \tfrac{1}{2}(112^\circ) = 56^\circ

The second bisector halves \angle DBC:

m\angle DBE = m\angle EBC = \tfrac{1}{2}(56^\circ) = 28^\circ

\overrightarrow{BD} lies inside \angle ABE, so by angle addition

\begin{aligned} m\angle ABE &= m\angle ABD + m\angle DBE \\ &= 56^\circ + 28^\circ = 84^\circ \end{aligned}

and m\angle EBC = 28^\circ.

Check. 84 + 28 = 112. Two bisections cut the angle into quarters: \overrightarrow{BE} sits three quarters of the way round from \overrightarrow{BA}, and 84 is three quarters of 112.

Example 12

Bisecting both angles of a linear pair

In Figure 13, A, O and B lie on one straight line and m\angle AOC = 130^\circ. \overrightarrow{OD} bisects \angle AOC and \overrightarrow{OE} bisects \angle COB. Find m\angle DOE.

Figure 13A linear pair, each angle cut in half by its own bisector.

Solution

\angle AOC and \angle COB form a linear pair, so they add to 180^\circ:

m\angle COB = 180^\circ - 130^\circ = 50^\circ

Each bisector takes half of its own angle:

\begin{aligned} m\angle DOC &= \tfrac{1}{2}(130^\circ) = 65^\circ \\ m\angle COE &= \tfrac{1}{2}(50^\circ) = 25^\circ \end{aligned}

\overrightarrow{OC} lies between \overrightarrow{OD} and \overrightarrow{OE}, so by angle addition

m\angle DOE = 65^\circ + 25^\circ = 90^\circ

The bisectors are perpendicular. Try it with 100^\circ instead of 130^\circ and the halves become 50^\circ and 40^\circ: still 90^\circ. That is the result on bisectors of a linear pair, seen in numbers before it is seen in general.

Example 13

A bisector inside a triangle

In triangle ABC of Figure 14, m\angle A = 50^\circ and m\angle B = 70^\circ. The bisector of \angle C meets \overline{AB} at D. Find m\angle ACD, m\angle ADC and m\angle BDC.

Figure 14\overline{CD} bisects \angle C and meets the opposite side at D.

Solution

First find the angle being bisected. The angles of a triangle add to 180^\circ (the triangle angle sum):

m\angle C = 180^\circ - 50^\circ - 70^\circ = 60^\circ

The bisector halves it:

m\angle ACD = m\angle DCB = \tfrac{1}{2}(60^\circ) = 30^\circ

Now look at the smaller triangle ADC. Its angles are 50^\circ at A, 30^\circ at C, and the unknown at D:

m\angle ADC = 180^\circ - 50^\circ - 30^\circ = 100^\circ

In triangle BDC the angles are 70^\circ at B and 30^\circ at C:

m\angle BDC = 180^\circ - 70^\circ - 30^\circ = 80^\circ

Check. \angle ADC and \angle BDC sit along the straight line \overline{AB}, so they form a linear pair: 100 + 80 = 180. They do.

Notice that the bisector reaches \overline{AB} closer to B than to A. It halves the angle, not the opposite side, and those are different things.

Boundaries

Congruence is about measure only. Position, arm length, the direction an angle faces and the letters used to name it all play no part. \angle ABC and \angle CBA are two names for one angle.

A figure cannot prove congruence by its looks, and a protractor cannot prove it either. A protractor reading is good to about a degree, so two angles read as 47^\circ and 47^\circ might really be 46.8^\circ and 47.3^\circ. Congruence is exact. It comes from given information, from matching marks, or from a reason such as vertical angles or a bisector.

\cong and = are not interchangeable. Congruence is a statement about two angles; equality of measures is a statement about two numbers. Every step of a proof keeps them apart, even though each implies the other.

The ray has to be inside. Suppose m\angle AOC = 70^\circ and \overrightarrow{OB} lies on the far side of \overrightarrow{OA}, with m\angle AOB = 40^\circ. Then m\angle BOC = 110^\circ, and 40 + 110 is not 70. The postulate still says something true here, but about a different angle: now A lies in the interior of \angle BOC, so m\angle BOA + m\angle AOC = m\angle BOC. Of three rays from one point that fit inside a half turn, it is the middle ray that splits, and the whole is the widest of the three angles.

The same condition is what stops a backwards ray being a bisector. Extend the bisector of a 60^\circ angle backwards past the vertex. The ray pointing the other way also makes equal angles with the two sides, 150^\circ with each, but it is outside the angle and the two angles it makes are not parts of \angle ABC, so it does not bisect.

The rays must share a vertex. m\angle AOB + m\angle BOC means nothing as a single angle unless both angles have their vertex at the same point O and share the side \overrightarrow{OB}.

The angles here are between 0^\circ and 180^\circ. A geometric angle is the opening between two rays. In trigonometry an angle can also be a directed turn, where 30^\circ, -30^\circ and 390^\circ are all different turns even though the first and last end in the same place. Congruence as defined here does not apply to turns of that kind. The same limit binds addition: two adjacent angles of 120^\circ and 100^\circ sit side by side, but together they sweep more than a half turn, and the angle between their outer sides is measured the short way round, at 360^\circ - 220^\circ = 140^\circ. Once the pieces add past 180^\circ, the postulate in this form no longer describes the figure; a signed measure that wraps round at 360^\circ is the tool that does.

Straight and reflex angles need their own wording. The interior above is defined only for an angle smaller than a straight angle. For a straight angle every ray from O off the line would count as inside, on one side or the other, so the postulate as stated does not cover it. The two pieces still add to 180^\circ, but that fact is stated separately as the linear pair rule. A straight angle likewise has two candidate bisectors, the ray perpendicular to the line at the vertex on either side, each making 90^\circ with both sides. And if you measure the reflex angle, the long way round from one side to the other, its bisector is exactly the backwards ray above. Which ray bisects depends on which of the two angles you mean.

An angle bisector is not a segment bisector. In a triangle the bisector of one angle meets the opposite side, but in general not at its midpoint, as Example 13 shows. It reaches that side at the midpoint only when the two sides forming the angle are equal. How the side is split is the triangle angle bisector theorem. A line through a midpoint is a different idea, the segment bisector.

A bisecting line bisects two angles. A full line through B along the bisector also runs through the angle vertical to \angle ABC, and it bisects that one too.

Where this goes next

Most of the angle facts that follow are statements that two angles are congruent. Vertical angles are congruent. The base angles of an isosceles triangle are congruent. When a transversal crosses parallel lines, corresponding angles are congruent. Each one is a reason to write \cong without measuring.

The addition postulate is the grammar of every angle calculation that follows. Put a ray inside a right angle and the pieces are complementary; inside a straight angle and they form a linear pair. In angle chases it is the step that says “this angle is those two together”. In proofs about angle pairs it is usually the line with the reason “Angle Addition Postulate”, and the proof of the triangle angle sum uses it to show that three angles at one point fill a straight line.

Congruent angles can also be made. With a compass and straightedge you can copy an angle anywhere on the page without knowing its measure, and bisecting an angle with compass and straightedge draws the bisector exactly, using the fold picture: two points equally far from the vertex, then the line of symmetry between them. The symmetry pays off again in the angle bisector distance theorem: a point lies on the bisector exactly when it is the same distance from both sides. In a triangle the three angle bisectors always meet at a single point, the incenter, which is the center of the circle that touches all three sides.

And the slide-and-turn argument above is the first case of a much bigger idea: two figures are congruent when rigid motions carry one onto the other. For whole triangles, congruent angles are half of what the congruence criteria check.

Exercises

Reading the definitions

Five angles measure m\angle A = 35^\circ, m\angle B = 55^\circ, m\angle C = 35^\circ, m\angle D = 145^\circ and m\angle E = 55^\circ. List every pair of congruent angles.
True or false: an angle whose arms are drawn 2 cm long cannot be congruent to an angle whose arms are drawn 9 cm long. Explain.
Write “angle JKL has the same measure as angle MNP” in two ways, once with \cong and once with =.
\angle ABC \cong \angle DEF. Is \angle CBA \cong \angle FED? Is \angle ABC \cong \angle FED? Explain.
In Figure 2, name the two angles whose measures add to m\angle AOC and the ray they share. Then write the angle addition postulate for the figure as an equation, and rewrite it to give m\angle BOC.
D lies in the interior of \angle ABC, with m\angle ABC = 70^\circ and m\angle ABD = 34^\circ. Does \overrightarrow{BD} bisect \angle ABC? Explain with numbers.

Finding measures

B is in the interior of \angle AOC. (a) If m\angle AOB = 28^\circ and m\angle BOC = 64^\circ, find m\angle AOC. (b) If instead m\angle AOC = 150^\circ and m\angle AOB = 87^\circ, find m\angle BOC.
\angle PQR \cong \angle XYZ and m\angle PQR = 73^\circ. Find m\angle XYZ, and the measure of each half when \angle XYZ is bisected.
\overrightarrow{BD} bisects \angle ABC and m\angle DBC = 37^\circ. Find m\angle ABC.
\angle 1 \cong \angle 2 and \angle 1 is a right angle. A ray inside \angle 2 splits it into two parts, one of 58^\circ. Find m\angle 2 and the other part.
Rays \overrightarrow{OB} and \overrightarrow{OC} lie inside \angle AOD, in the order OA, OB, OC, OD. If m\angle AOB = 25^\circ, m\angle BOC = 40^\circ and m\angle COD = 55^\circ, find m\angle AOD.
\angle A \cong \angle B, and the two angles are complementary. Find the measure of each.
\angle 1 \cong \angle 2, and \angle 1 is supplementary to an angle of 128^\circ. Find m\angle 2.
At 2 o’clock and at 10 o’clock, the hour and minute hands of a clock make an angle. Find the measure of each angle and say whether they are congruent.

Using algebra

B is in the interior of \angle AOC, m\angle AOB = (x + 12)^\circ, m\angle BOC = (2x)^\circ and m\angle AOC = 78^\circ. Find x and both pieces.
B is in the interior of \angle AOC, m\angle AOB = (4x + 3)^\circ, m\angle BOC = (x + 17)^\circ and m\angle AOC = (6x - 5)^\circ. Find x and all three measures.
B is in the interior of \angle AOC, m\angle AOC = 111^\circ, and m\angle AOB is twice m\angle BOC. Find both pieces.
\overrightarrow{BD} bisects \angle ABC, with m\angle ABD = (4x - 6)^\circ and m\angle DBC = (2x + 18)^\circ. Find x, each half, and m\angle ABC.
\overrightarrow{BD} bisects \angle ABC, with m\angle ABD = (x + 15)^\circ and m\angle ABC = (3x - 5)^\circ. Find x, m\angle ABD and m\angle ABC.
Two congruent angles measure \displaystyle \left(\frac{x}{2} + 20\right)^\circ and (x - 15)^\circ. Find x and the measure of each angle.
\angle 1 \cong \angle 2, and \angle 1 is supplementary to \angle 3. If m\angle 2 = (x + 30)^\circ and m\angle 3 = (4x)^\circ, find x and all three measures.
\angle P \cong \angle Q and \angle Q \cong \angle R, with m\angle P = (2x + 10)^\circ and m\angle R = (4x - 20)^\circ. Find x and m\angle Q.
B is in the interior of \angle AOC, m\angle AOB = (x^2)^\circ, m\angle BOC = (3x)^\circ and m\angle AOC = 108^\circ. Find every value of x that gives a real figure, and the pieces it gives.
\overrightarrow{BD} bisects \angle ABC, with m\angle ABD = (x^2)^\circ and m\angle DBC = (2x + 48)^\circ. Find every value of x that works, and the measure of \angle ABC that each one gives.

Several rays and repeated bisection

Exercises 25 and 26 use four rays in the order \overrightarrow{OA}, \overrightarrow{OB}, \overrightarrow{OC}, \overrightarrow{OD}, as in Figure 12.

m\angle AOC = 70^\circ, m\angle BOD = 80^\circ and m\angle AOD = 115^\circ. Find m\angle BOC, then m\angle AOB and m\angle COD.
m\angle AOB = 20^\circ, m\angle AOC = 65^\circ and m\angle AOD = 130^\circ. Find m\angle BOC, m\angle COD and m\angle BOD.
m\angle ABC = 120^\circ. \overrightarrow{BD} bisects \angle ABC and \overrightarrow{BE} bisects \angle ABD. Find m\angle EBC.
\overrightarrow{BD} bisects \angle ABC and \overrightarrow{BE} bisects \angle ABD. If m\angle ABE = 17^\circ, find m\angle ABC.
\angle AOC and \angle COB form a linear pair with m\angle AOC = 48^\circ. \overrightarrow{OD} bisects \angle AOC and \overrightarrow{OE} bisects \angle COB. Find m\angle DOE and m\angle DOB.
Two lines cross, and one of the four angles measures 70^\circ. Find the angle between the bisector of that angle and the bisector of the angle vertical to it. What does your answer say about the two bisectors?

Bisectors in triangles

Triangle ABC is equilateral, and the bisector of \angle A meets \overline{BC} at D. Find m\angle BAD and m\angle ADB.
In triangle ABC, the bisector of \angle A makes an angle of 25^\circ with side \overline{AB}, and m\angle B = 65^\circ. Find m\angle C.

Reasoning

Explain why any two right angles are congruent.
\angle 1 \cong \angle 2 and \angle 2 \cong \angle 3. Must \angle 1 \cong \angle 3? Name the property you are using.
Two angles are supplementary and congruent. Show that both are right angles.
A student measures two angles with a protractor, gets 47^\circ both times, and writes that the angles are congruent. Is that justified? Explain.
Explain why a slide followed by a turn is always enough to carry an angle onto any angle congruent to it, with no flip needed.
Explain why, if B is in the interior of \angle AOC, the angle \angle AOB must be smaller than \angle AOC.
m\angle AOB = 40^\circ and m\angle AOC = 70^\circ, but \overrightarrow{OB} and \overrightarrow{OC} lie on opposite sides of line OA, so that A is in the interior of \angle BOC. Explain why m\angle AOB + m\angle BOC = m\angle AOC is false here, and find m\angle BOC.
Rays \overrightarrow{OB} and \overrightarrow{OC} lie inside \angle AOD in the order OA, OB, OC, OD, and m\angle AOB = m\angle COD. Prove that m\angle AOC = m\angle BOD.
Ray \overrightarrow{OB} lies inside a right angle \angle AOC. Prove that \angle AOB and \angle BOC are complementary.
\angle 1 and \angle 2 are complementary, \angle 3 and \angle 4 are complementary, and \angle 1 \cong \angle 3. Prove that \angle 2 \cong \angle 4. Then find m\angle 4 when m\angle 1 = 27^\circ.
The segment addition postulate says that if B is between A and C on a line, then AB + BC = AC. Match each part of it to the matching part of the angle addition postulate.
m\angle ABC = 60^\circ and \overrightarrow{BD} is its bisector. \overrightarrow{BF} points in the direction opposite to \overrightarrow{BD}. Find m\angle ABF and m\angle CBF, and explain why \overrightarrow{BF} is not a bisector of \angle ABC even though the two angles are equal.
\angle ABC \cong \angle PQR. \overrightarrow{BD} bisects \angle ABC and \overrightarrow{QS} bisects \angle PQR. Show that all four of \angle ABD, \angle DBC, \angle PQS and \angle SQR are congruent.
Explain why both halves made by any angle bisector are acute angles.
A, B and C lie on one line, with B between A and C, so \angle ABC is a straight angle. Describe its bisector and explain why.
\overrightarrow{BD} bisects \angle ABC and \overrightarrow{BE} bisects \angle ABD. Show that m\angle EBC = 3\, m\angle ABE, whatever the size of \angle ABC.

Answers

\angle A \cong \angle C and \angle B \cong \angle E. Those are the only equal measures. \angle D is supplementary to \angle A and to \angle C, but not congruent to either.
False. Arm length plays no part in an angle’s measure, so the two angles are congruent whenever their openings are equal, however long the arms are drawn.
\angle JKL \cong \angle MNP, and m\angle JKL = m\angle MNP. The first compares the angles, the second their measures.
Yes to both. \angle CBA is the same angle as \angle ABC and \angle FED is the same angle as \angle DEF; only the order of the outer letters differs. The vertex in the middle is what names the angle.
\angle AOB and \angle BOC, which share the ray \overrightarrow{OB} running inside \angle AOC. The postulate reads m\angle AOB + m\angle BOC = m\angle AOC, and subtracting m\angle AOB from both sides gives m\angle BOC = m\angle AOC - m\angle AOB.
No. By angle addition m\angle DBC = 70^\circ - 34^\circ = 36^\circ, and 36^\circ \ne 34^\circ, so the two parts are not congruent. The bisector would make 35^\circ with each side.
(a) 92^\circ. The pieces add to the whole: 28^\circ + 64^\circ = 92^\circ. (b) 63^\circ. The missing piece is the whole minus the known piece: 150^\circ - 87^\circ = 63^\circ.
73^\circ, and each half measures 36.5^\circ. Congruent angles have equal measures, and a bisector splits 73^\circ into two equal parts, 73 \div 2 = 36.5. A bisector need not give a whole number.
m\angle ABC = 74^\circ. The whole is two equal halves: 2(37^\circ).
m\angle 2 = 90^\circ and the other part is 32^\circ. A right angle measures 90^\circ, and a congruent angle has the same measure. The two parts fill \angle 2, so the other is 90^\circ - 58^\circ = 32^\circ.
120^\circ. Applying the postulate twice, 25^\circ + 40^\circ + 55^\circ = 120^\circ.
45^\circ each. Equal measures adding to 90^\circ give 2m = 90, so m = 45.
52^\circ. m\angle 1 = 180^\circ - 128^\circ = 52^\circ, and \angle 2 has the same measure.
60^\circ both times, so the angles are congruent. The hour marks are 360^\circ \div 12 = 30^\circ apart; at 2 o’clock the hands are two marks apart, 2(30) = 60, and at 10 o’clock they are also two marks apart, the other way round.
x = 22, with pieces 34^\circ and 44^\circ. From (x + 12) + 2x = 78, 3x = 66. Check: 34 + 44 = 78.
x = 25, and the measures are m\angle AOB = 103^\circ, m\angle BOC = 42^\circ, m\angle AOC = 145^\circ. From (4x + 3) + (x + 17) = 6x - 5, 5x + 20 = 6x - 5. Check: 103 + 42 = 145.
m\angle AOB = 74^\circ and m\angle BOC = 37^\circ. Writing m\angle BOC = y, the postulate gives 2y + y = 111. So 3y = 111 and y = 37.
x = 12, each half is 42^\circ and m\angle ABC = 84^\circ. The halves are equal: 4x - 6 = 2x + 18, so 2x = 24. Check: 4(12) - 6 = 42 and 2(12) + 18 = 42.
x = 35, m\angle ABD = 50^\circ and m\angle ABC = 100^\circ. The whole is twice the half: 2(x + 15) = 3x - 5, so 2x + 30 = 3x - 5 and x = 35.
x = 70, and each angle is 55^\circ. From \displaystyle \frac{x}{2} + 20 = x - 15, \displaystyle 35 = \frac{x}{2}. Check: 35 + 20 = 55 and 70 - 15 = 55.
x = 30, with m\angle 1 = m\angle 2 = 60^\circ and m\angle 3 = 120^\circ. Since m\angle 1 = m\angle 2 = x + 30 and \angle 1 is supplementary to \angle 3, (x + 30) + 4x = 180, so 5x = 150.
x = 15 and m\angle Q = 40^\circ. By transitivity \angle P \cong \angle R, so 2x + 10 = 4x - 20 and x = 15. Then m\angle P = 40^\circ, and \angle Q has the same measure.
Only x = 9, giving 81^\circ and 27^\circ. The postulate gives x^2 + 3x = 108, so x^2 + 3x - 108 = 0 and (x - 9)(x + 12) = 0. The root x = -12 would make m\angle BOC = 3(-12) = -36^\circ, and no angle has a negative measure, so it is rejected.
x = 8, giving m\angle ABC = 128^\circ, or x = -6, giving m\angle ABC = 72^\circ. Setting the halves equal, x^2 = 2x + 48, so x^2 - 2x - 48 = 0 and (x - 8)(x + 6) = 0. For x = 8 each half is 64^\circ; for x = -6 each half is 36^\circ. Both halves are positive in both cases, so both values work, unlike Exercise 23.
m\angle BOC = 35^\circ, m\angle AOB = 35^\circ, m\angle COD = 45^\circ. Adding the two overlapping angles counts \angle BOC twice: m\angle AOC + m\angle BOD = m\angle AOD + m\angle BOC, so m\angle BOC = 70^\circ + 80^\circ - 115^\circ = 35^\circ. Then m\angle AOB = 70^\circ - 35^\circ and m\angle COD = 80^\circ - 35^\circ.
m\angle BOC = 45^\circ, m\angle COD = 65^\circ, m\angle BOD = 110^\circ. Each is a difference of angles measured from \overrightarrow{OA}: 65^\circ - 20^\circ = 45^\circ, 130^\circ - 65^\circ = 65^\circ and 130^\circ - 20^\circ = 110^\circ.
m\angle EBC = 90^\circ. m\angle ABD = m\angle DBC = 60^\circ, then m\angle EBD = 30^\circ, and m\angle EBC = 30^\circ + 60^\circ.
m\angle ABC = 68^\circ. \angle ABE is half of \angle ABD, which is half of \angle ABC, so m\angle ABC = 4(17^\circ).
m\angle DOE = 90^\circ and m\angle DOB = 156^\circ. m\angle COB = 180^\circ - 48^\circ = 132^\circ, so the halves are 24^\circ and 66^\circ, and 24^\circ + 66^\circ = 90^\circ. Also m\angle AOD = 24^\circ, and \angle AOD and \angle DOB form a linear pair, so m\angle DOB = 180^\circ - 24^\circ.
180^\circ: the two bisectors point in opposite directions and together make one straight line. Going round from one bisector to the other passes half of the 70^\circ angle, a whole 110^\circ neighbor, and half of the vertical 70^\circ angle: 35^\circ + 110^\circ + 35^\circ = 180^\circ.
m\angle BAD = 30^\circ and m\angle ADB = 90^\circ. Each angle of an equilateral triangle is 60^\circ, so the half is 30^\circ; in triangle ABD, m\angle ADB = 180^\circ - 60^\circ - 30^\circ. The bisector is perpendicular to the opposite side.
m\angle C = 65^\circ. The bisector makes 25^\circ with one side, so m\angle A = 50^\circ, and m\angle C = 180^\circ - 50^\circ - 65^\circ.
Every right angle measures exactly 90^\circ by definition, so any two right angles have equal measures, which is what congruent means.
Yes, by the transitive property. m\angle 1 = m\angle 2 and m\angle 2 = m\angle 3 give m\angle 1 = m\angle 3.
Call the common measure m. Supplementary means the measures add to 180^\circ, and congruent means they are equal, so m + m = 180 and m = 90. Both are right angles.
No. A protractor is only good to about a degree, so the true measures might differ by a fraction of a degree. The reading suggests congruence; a reason from the figure or the given information is needed to prove it.
Slide the vertex of one onto the other’s vertex, then turn one arm onto the other angle’s first arm. The second arm then lies either on the other angle’s second arm, and you are done, or on its mirror image. In that case turn the other arm onto the first arm instead; the arms swap roles and both land. An angle is just its two rays, neither one first, so this always works.
By the postulate, m\angle AOC = m\angle AOB + m\angle BOC. The measure m\angle BOC is positive, since \overrightarrow{OB} and \overrightarrow{OC} are different rays, so m\angle AOC is larger than m\angle AOB.
m\angle BOC = 110^\circ. The postulate only splits an angle at a ray inside it, and \overrightarrow{OB} is not inside \angle AOC. The ray inside is \overrightarrow{OA}, so the true statement is m\angle BOA + m\angle AOC = m\angle BOC, giving 40^\circ + 70^\circ = 110^\circ. The false sum would claim 40 + 110 = 70.
B is inside \angle AOC, so m\angle AOC = m\angle AOB + m\angle BOC. C is inside \angle BOD, so m\angle BOD = m\angle BOC + m\angle COD. Since m\angle AOB = m\angle COD, the two right sides are equal, so m\angle AOC = m\angle BOD. Adding the same piece to two equal pieces gives equal wholes, the reverse of Example 10.
By the postulate, m\angle AOB + m\angle BOC = m\angle AOC = 90^\circ. Two angles whose measures add to 90^\circ are complementary by definition.
Complementary angles add to 90^\circ, so m\angle 2 = 90^\circ - m\angle 1 and m\angle 4 = 90^\circ - m\angle 3. Since \angle 1 \cong \angle 3, the measures m\angle 1 and m\angle 3 are equal, so the right-hand sides are equal, and m\angle 2 = m\angle 4, that is, \angle 2 \cong \angle 4. For the numbers, m\angle 3 = 27^\circ too, so m\angle 4 = 90^\circ - 27^\circ = 63^\circ.
The point B between A and C matches the point B in the interior of \angle AOC. The lengths AB and BC match the measures m\angle AOB and m\angle BOC, and the whole length AC matches m\angle AOC. In both, the pieces made by something in the middle add to the whole.
m\angle ABF = m\angle CBF = 150^\circ. \overrightarrow{BD} makes 30^\circ with each side, and \angle ABD, \angle ABF form a linear pair, so m\angle ABF = 180^\circ - 30^\circ; the same for \angle CBF. \overrightarrow{BF} is not a bisector because F is not in the interior of \angle ABC, and the two 150^\circ angles are not halves of the 60^\circ angle.
Congruent angles have equal measures; call it a. Each bisector cuts its angle into two halves of \tfrac{a}{2}, so all four angles measure \tfrac{a}{2} and are congruent.
An angle here measures less than 180^\circ, so each half measures less than \tfrac{1}{2}(180^\circ) = 90^\circ, which makes it acute. (For a straight angle each half is exactly 90^\circ; see Exercise 47.)
It is the ray from B perpendicular to the line AC, on either side of it, since a straight angle has no single inside (see Boundaries). The two halves must add to 180^\circ and be equal, so each is 90^\circ.
Let m\angle ABE = t. \overrightarrow{BE} bisects \angle ABD, so m\angle ABD = 2t, and \overrightarrow{BD} bisects \angle ABC, so m\angle ABC = 4t. By angle addition m\angle EBC = 4t - t = 3t.