Geometry

Given, prove and the diagram

A theorem is a sentence about every triangle, or every pair of lines. How do you turn it into something a proof can start from, and which of the things you can see in its figure are facts?

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Given, prove and the diagram

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Why this exists

Theorems are usually stated in words: “the base angles of an isosceles triangle are congruent”, “every point on the perpendicular bisector of a segment is equally far from its ends”. Words like these speak about every triangle of a kind at once, and that is their point. But a proof cannot work on the base angles of no triangle in particular. Each step has to point: this angle, that segment, the place where these two lines meet.

So before its first step, every proof sorts the sentence into what may be assumed and what must be shown, names the figure with letters, and draws it. Euclid did this in every proposition. In Book I, Proposition 15 he states the result in words, then writes “Let the straight lines AB and CD cut one another at the point E”, then “I say that the angle CEA equals the angle DEB”, and only then argues. Proclus, writing on Euclid in the fifth century, named these two steps; in English they are the setting-out and the specification. School geometry calls them the Given and the Prove.

The drawing brings a problem of its own. A proof is a chain of reasons, each one a definition, postulate or theorem or something you were given, and a figure shows far more than the words do. A segment in it has a length you could measure, two lines in it meet at an angle you could read with a protractor, and two lines that do not meet look as if they never will. If every one of those appearances counted as a fact, a proof could stop at “look”. If none counted, you could not even say which point lies on which line without writing it out. Geometry settles this with a rule about what a figure is allowed to say, and every proof you write leans on it.

Definition

The set-up: Given, Prove and the diagram

A theorem proved from a figure has the shape “if p, then q”, even when the words hide it (see conditionals). Its set-up has three parts.

The Given is the hypothesis p, written about named points. It lists everything the proof may assume about the figure, and nothing else.
The Prove is the conclusion q, written about the same named points. It is the one statement the proof must end on.
The diagram is a figure drawn to fit the Given and labeled with the same letters. Each measure fact in the Given (a fact about size, such as equal lengths, a right angle or parallel lines, sorted out exactly in the next subsection) is drawn as a mark; the Prove is never marked.

On this site a set-up is written as two lines, several givens separated by semicolons, then the diagram, then often a Plan: a sentence or two naming the route the proof will take. Take “the segment joining the midpoints of two sides of a triangle is parallel to the third side and half as long”. Its set-up is

Given. \triangle ABC; D is the midpoint of \overline{AB}; E is the midpoint of \overline{AC}.

Prove. \overline{DE} \parallel \overline{BC} and DE = \tfrac{1}{2}BC.

Figure 1The diagram for the midsegment theorem. The ticks say what is given; nothing says what is to be proved.

The Given names the triangle and both midpoints, because the words mention them. Being a midpoint is a measure fact, so Figure 1 carries ticks: one on each half of \overline{AB}, two on each half of \overline{AC}. Nothing marks \overline{DE} and \overline{BC} as parallel and no length is written on either, since that is the Prove and it has to be earned. The proof itself belongs to the triangle midsegment theorem.

What a diagram lets you assume

From a figure you may read where things are:

incidence: which points lie on which lines, segments, rays and circles, and where lines meet
collinearity: which points lie on one line, the word from collinear and coplanar points
betweenness: the order of points along a line, so that D lies between B and C
adjacency and position: which angles share a side, which side of a line a point is on, and whether a point or ray is inside an angle or a triangle

From a figure you may never read how big things are. Equal lengths, equal angles, measures, right angles, perpendicular lines, parallel lines, midpoints and bisectors, and which of two segments is longer: each must be given, in the words of the problem or by a mark on the figure.

The first kind of fact is about position; the second is about measure. Figure 2 shows one figure two ways. On the left is what it says: triangle ABC, a point D on \overline{BC}, and \overline{AD}. On the right, drawn dashed, is what the eye adds: that \overline{AD} is square to \overline{BC}, that D is the middle of \overline{BC}, and that the triangle’s two slanting sides match.

Figure 2Left, what the figure says. Right, what the eye adds. None of the dashed marks is in the figure.

From the left panel you may read that B, D and C are collinear with D between B and C, that \angle ADB and \angle ADC sit side by side along \overline{BC}, and that ray \overrightarrow{AD} lies inside \angle BAC. You may not read any of the dashed marks.

Intuition

The letters are a choice of one triangle to stand for all of them. \triangle ABC means “a triangle, any one, with D and E where the words put them”. Whatever the proof shows about it using nothing else, it has shown about every triangle.

That is why the diagram is drawn general, not special. If the words say “a triangle”, draw one with no two sides equal and no right angle. A special drawing shows facts the Given does not contain, and the eye reads them in, which is exactly what the rule for what a diagram lets you assume forbids. The names follow the words: vertices in order around a figure, and a letter of its own for any point the words only describe.

Put the other way round: a figure is a sketch of one figure that fits the given, but a proof has to hold for every figure that fits. If all you are told is that D lies on \overline{BC}, then each triangle in Figure 3 fits as well as Figure 2 does.

Figure 3Three figures that fit “D lies on \overline{BC}”. A proof from that given has to work for every one of them.

In all three, D is between B and C and the two angles at D make a straight line: those are facts about position, and every figure that fits shares them. But \overline{AD} is square to the base in only one of them, and D is at the middle in only one. A reason that uses either of those proves something about one drawing, not about the figure the problem describes.

Even a careful drawing cannot certify a measure. Figure 2 was in fact drawn with \overline{AD} about 1^\circ away from square, which no eye can see. Two lengths that agree to the width of a pencil line may differ by a thousandth of a millimeter, and that is still different.

Results

From statement to set-up

In the theorem	In the set-up
the hypothesis, the if-part	the Given
the conclusion, the then-part	the Prove
“a triangle”, “a point”, “two lines”	named objects: \triangle ABC, C, \ell and m
a measure fact in the hypothesis	a Given line and a mark on the diagram
a position fact in the hypothesis	a Given line, and the diagram drawn to show it
the conclusion	no mark anywhere

A proof of the set-up proves the theorem. If a proof uses nothing about the named figure beyond the Given, it holds for every figure the words describe.

Swapping the Given and the Prove gives the converse. The converse needs its own set-up and its own proof, and it may be false.

What a diagram may be used for

The rule. A figure may be used for positions: incidence, collinearity, betweenness, adjacency, inside and outside. Every measure fact, including congruence, perpendicularity and parallelism, must be given in words or by a mark, or proved.

Marks carry measure facts. These marks are part of what a figure says, as much as the words are:

Mark	What it says
matching ticks on segments	the segments are congruent: equal lengths
matching arcs in angles	the angles are congruent: equal measures
a small square in a corner	the angle is a right angle, so the lines are perpendicular
matching arrowheads on lines	the lines are parallel
a number or expression	that length or measure

Figure 4 is Figure 2 with marks. From it you may now use BD = DC and \overline{AD} \perp \overline{BC}, because the figure says them.

Figure 4The same figure with marks. Now BD = DC and \overline{AD} \perp \overline{BC} are part of what it says.

What cannot be read can often be proved. In Figure 4, AB = AC is not marked, but it follows from the marks, since A is on the perpendicular bisector of \overline{BC} (the perpendicular bisector theorem). The rule limits what you may look at, not what you may prove.

Why it is true

Why one named figure stands for all of them

The theorem in words says: for every figure, if p then q. The way to prove a claim about every figure is to take one about which nothing is known except p, and show q. Naming it \triangle ABC is how you take one. The letters carry no information of their own, so the same lines, word for word, work for any other triangle that fits p. In logic this step is called universal generalization (see the universal quantifier).

The Prove is never marked or given, because an argument that assumes q to reach q shows nothing. The converse needs its own proof because q \Rightarrow p is a different statement from p \Rightarrow q, and either can be false while the other is true (a theorem and its converse).

Why positions may be read and measures may not

The rule about the diagram is a convention, so the question is why this line is the right place to draw it.

Positions are built in or stable. A point placed on a line stays on it however the figure is redrawn, because it was placed there on purpose. A point strictly between two others on a line stays between them when each is nudged a little along the line. Measures are the opposite: a nudge too small to see breaks BD = DC, turns 90^\circ into 89.9^\circ, and makes two parallel lines meet a kilometer off the page. Kenneth Manders named the two kinds co-exact (position, stable under small changes) and exact (measure, destroyed by them), and observed that Euclid’s proofs take only co-exact facts from their diagrams.

Positions are hard to write and easy to see. Euclid never stated why a line entering a triangle through one side must leave through another; he read it off the picture. Moritz Pasch first wrote axioms of betweenness in 1882, and in 2009 Jeremy Avigad, Edward Dean and John Mumma showed that Euclid’s way of reading position from a diagram can be written down as a precise formal system. So reading position is safe: it can always be backed by axioms, and the convention spares you from writing them out.

Worked examples

Example 1

Reading a figure

For each statement, say whether it may be read from the left panel of Figure 2, follows from what may be read, or neither: (a) D lies between B and C; (b) \overline{AD} \perp \overline{BC}; (c) m\angle ADB + m\angle ADC = 180^\circ; (d) AB = AC.

Solution

(a) Read. The order of points along a line is betweenness, a position fact.

(b) Neither. Perpendicular is a measure fact, and there is no square at D.

(c) Follows. The sum is a measure, so it is not read, but it is proved from what is read: B, D and C are collinear with D between, so \overrightarrow{DB} and \overrightarrow{DC} are opposite rays, and \angle ADB and \angle ADC form a linear pair. Linear pairs add to 180^\circ.

(d) Neither. Equal lengths are a measure fact, and \overline{AB} and \overline{AC} carry no ticks.

Example 2

A square that is not marked

In Figure 5, A, B and C lie on a line and m\angle ABD = 34^\circ. Find m\angle DBC, and say whether m\angle EBC can be found.

Figure 5\angle DBE looks square. Nothing in the figure says so.

Solution

The figure shows B between A and C, so \overrightarrow{BA} and \overrightarrow{BC} are opposite rays and \angle ABD and \angle DBC form a linear pair:

m\angle DBC = 180^\circ - 34^\circ = 146^\circ

For m\angle EBC you would need m\angle DBE. It looks like 90^\circ, but it carries no square, so it is unknown and m\angle EBC cannot be found. The position facts still hold: \overrightarrow{BE} is inside \angle DBC, so m\angle DBE + m\angle EBC = 146^\circ by the angle addition postulate.

Had the figure marked \angle DBE as a right angle, m\angle EBC = 146^\circ - 90^\circ = 56^\circ.

Example 3

Using the marks

In Figure 4, BD = 3x - 1 and DC = x + 7. Find x and BC, and find m\angle ADC.

Solution

The ticks say BD = DC, so

\begin{aligned} 3x - 1 &= x + 7 \\ 2x &= 8 \\ x &= 4 \end{aligned}

and each piece is 3(4) - 1 = 11. The figure shows D between B and C, a position fact, so the segment addition postulate applies:

BC = BD + DC = 11 + 11 = 22

The square at D says m\angle ADB = 90^\circ. \angle ADB and \angle ADC form a linear pair, so m\angle ADC = 180^\circ - 90^\circ = 90^\circ.

Each step used a mark for a measure and the picture only for a position.

Example 4

An angle that looks right

In a figure, \overleftrightarrow{AC} and \overleftrightarrow{BD} cross at E, and \angle AEB looks like a right angle. Its measure is (4x + 8)^\circ, and m\angle CED = (6x - 32)^\circ. Find m\angle AEB and m\angle AED.

Solution

Reading the figure: E is between A and C and between B and D, so \angle AEB and \angle CED are vertical angles, and equal:

\begin{aligned} 4x + 8 &= 6x - 32 \\ 40 &= 2x \\ x &= 20 \end{aligned}

So m\angle AEB = 4(20) + 8 = 88^\circ, and its neighbor \angle AED is 180^\circ - 88^\circ = 92^\circ. The angle that looked right is not.

Example 5

Lines that look parallel

In Figure 6, find x. Then find it again when the problem adds “\ell \parallel m”.

Figure 6\ell and m look parallel, but the figure carries no arrowheads.

Solution

As drawn, x cannot be found. \ell and m look parallel, but parallel is a measure fact, and the figure has no arrowheads. The crossing with \ell gives its four angles, 65^\circ, 115^\circ, 65^\circ and 115^\circ, from vertical angles and linear pairs, and says nothing about the crossing with m. Figure 6 was in fact drawn with m tilted 2^\circ from \ell, so the angle drawn at x is 67^\circ.

Given \ell \parallel m, the two marked angles sit in matching positions at the two crossings, so they are corresponding angles and equal: x = 65.

Example 6

Two circles

In Figure 7, A is the center of one circle and B the center of the other. Show that \triangle ABC is equilateral.

Figure 7Circle A passes through B, circle B passes through A, and both pass through C.

Solution

Being a center is a measure fact (every point of the circle is the same distance from it), so it had to be given, and it was. The usual way to give it is the one used here, in words, or by naming the circle after its center: “circle A” means the circle whose center is A. A dot drawn near the middle of a circle does not say it, however central it looks. Everything else is incidence, read from the figure: B and C lie on the circle centered at A, and A and C lie on the circle centered at B.

AB and AC are radii of the circle centered at A, so AB = AC. BA and BC are radii of the circle centered at B, so BA = BC. Then AB = AC = BC, and the triangle is equilateral.

The two circles look the same size, but that was never read. It was proved: both radii equal AB.

Example 7

Two right angles

Set up the statement “all right angles are congruent”.

Solution

First find the if-then inside it: if two angles are right angles, then they are congruent. The words speak of two angles and say nothing about where they are, so they get separate names.

Given. \angle ABC and \angle DEF are right angles.

Prove. \angle ABC \cong \angle DEF.

Figure 8Two right angles, drawn apart, because the statement says nothing about where they sit.

Being a right angle is a measure fact, so each angle gets a square. Figure 8 draws the two apart and turned different ways. Drawing them with a shared vertex or side would add a position fact the Given does not contain.

Plan. Each angle measures 90^\circ by the definition of a right angle, and angles with equal measures are congruent. The proof with a reason on every line is in proving angle pair theorems.

Example 8

A statement not in if-then form

Set up “the acute angles of a right triangle are complementary”.

Solution

The hypothesis is hiding in the subject of the sentence: if a triangle is a right triangle, then its two acute angles are complementary. The triangle needs a name, and the right angle needs a corner. Put it at C.

Given. \triangle ABC with \angle C a right angle.

Prove. \angle A and \angle B are complementary.

Figure 9The Given fixes which corner is the right angle, so “the acute angles” become \angle A and \angle B.

Without fixing the corner of the right angle, “the acute angles” would have no letters. The Prove could equally be written m\angle A + m\angle B = 90^\circ, which is what complementary means and the form the proof’s last line will reach.

Plan. By the triangle angle sum, m\angle A + m\angle B + 90^\circ = 180^\circ; subtract 90^\circ from both sides.

Example 9

Opposite which side?

Set up “if two sides of a triangle are congruent, then the angles opposite those sides are congruent”.

Solution

Name the triangle \triangle ABC and choose \overline{AB} and \overline{AC} as the two congruent sides. The angle opposite \overline{AB} is at the vertex that side misses, \angle C. The angle opposite \overline{AC} is \angle B.

Given. \triangle ABC with \overline{AB} \cong \overline{AC}.

Prove. \angle B \cong \angle C.

Figure 10Ticks on the two given sides. The angles at B and C are what the proof must reach.

The choice of sides was free, since the words do not say which two. The two sides get ticks; the angles at B and C get no arcs. This is the Base Angles Theorem.

Plan. Both angles sit in one triangle, so there is no pair of triangles to compare. An added segment makes one: bisect \angle A, meeting \overline{BC} at D, show \triangle ABD \cong \triangle ACD, and match their angles at B and C. A segment added this way is an auxiliary line, and the full proof is proving the Base Angles Theorem.

Example 10

Any point on the line

Set up “every point on the perpendicular bisector of a segment is equally far from the segment’s endpoints”.

Solution

“Every point on” holds the hypothesis: if a point lies on the perpendicular bisector of a segment, then it is equally far from the endpoints. Name the segment \overline{AB}, the bisector \ell, their crossing point M, and the point C. “Perpendicular bisector” unpacks into two facts, a right angle and a midpoint, and each becomes a Given.

Given. \ell \perp \overline{AB} at M; M is the midpoint of \overline{AB}; C is a point on \ell.

Prove. CA = CB.

Figure 11C is any point of \ell. The figure shows one; the proof must work for all of them.

Both measure facts are marks: the square at M, and the ticks on \overline{AM} and \overline{MB}. \overline{CA} and \overline{CB} carry no ticks. Figure 11 puts C above M, but the Given says only that C is on \ell: it could be below \overline{AB}, or M itself, and the proof has to cover every place.

Plan. If C is M, then CA = MA = MB = CB. Otherwise \triangle AMC \cong \triangle BMC by SAS, one of the congruence criteria set out in a later lesson (AM = MB, right angles at M, shared side \overline{MC}), so CA = CB. This is the perpendicular bisector theorem.

Example 11

The converse

Write the converse of the statement in Example 10 and set it up.

Solution

The converse swaps the two parts: if a point is equally far from the endpoints of a segment, then it lies on the segment’s perpendicular bisector.

Given. \overline{AB}; C is a point with CA = CB.

Prove. C lies on the perpendicular bisector of \overline{AB}.

Figure 12The converse. The ticks on \overline{CA} and \overline{CB} are now given, and the square at M is gone.

The perpendicular bisector has moved into the Prove, so there is no \ell to name. The diagram builds a point instead: M, the midpoint of \overline{AB}. Every segment has one, and its ticks are allowed because M was made that way. With M named, the Prove reads: C is M, or \overleftrightarrow{CM} \perp \overline{AB}. Figure 12 differs from Figure 11 exactly where the Given and the Prove changed places: ticks on \overline{CA} and \overline{CB}, no square at M. The dashed \overline{CM} is added for the plan.

Plan. If C is M, there is nothing to prove. Otherwise \triangle AMC \cong \triangle BMC by SSS, another of those later criteria, so \angle AMC \cong \angle BMC; they form a linear pair, so each is 90^\circ. The theorem and its converse together make a biconditional.

Example 12

A point the statement only describes

Set up “the diagonals of a parallelogram bisect each other”.

Solution

The hypothesis is that a quadrilateral is a parallelogram. Name it ABCD, going around, and state the definition, both pairs of opposite sides parallel, since that is what the proof can use. The statement talks about where the diagonals cross without naming the place, so the set-up gives it a letter, E. “Bisect each other” unpacks into two equations, one for each diagonal.

Given. \overline{AB} \parallel \overline{DC}; \overline{AD} \parallel \overline{BC}; the diagonals \overline{AC} and \overline{BD} meet at E.

Prove. AE = EC and BE = ED.

Figure 13The crossing point has no name in the statement, so the set-up gives it one, E.

Arrowheads mark the two pairs of parallel sides, and the four pieces of the diagonals carry no ticks. That the diagonals cross inside the figure is a position fact, which the diagram may show.

Plan. Once the congruence criteria are proved, show \triangle ABE \cong \triangle CDE by ASA: AB = CD (opposite sides of a parallelogram), and the angles at each end are alternate interior angles along \overline{AB} \parallel \overline{DC}. Matching sides give both equations. The full proof is in proving properties of a parallelogram.

Boundaries

Not every theorem is one figure and one if-then. “Every segment has exactly one midpoint” claims that something exists and is unique, and “a triangle has at most one right angle” is usually proved by assuming the opposite (indirect proof). Those still get a Given and a Prove, but the Prove is an existence claim, or the proof adds an assumption it will contradict.

The set-up must say exactly what the words say. A Given with an extra fact proves only a special case. A Prove that drops part of the conclusion proves less than was claimed, and one that adds to it claims more than the theorem does.

One diagram, several cases. When the hypothesis allows different arrangements, one drawing shows only one of them. An altitude of an obtuse triangle can land outside the opposite side (altitudes), and C in Example 10 can be M. The proof has to handle each case, and it helps to draw one diagram for each.

The figure must be drawn to fit the given. Positions are read on trust that the sketch places points where the words put them. The well-known false proof that every triangle is isosceles draws the meeting point of an angle bisector and a perpendicular bisector inside the triangle, when for a triangle that is not isosceles it always lies outside, and reads the wrong betweenness from that. When a point is made by crossing two lines, where it lands is a claim to check, not something to look at.

Words beat the picture. If the given says something the figure seems to contradict, the given wins. “Not drawn to scale” changes nothing under this rule, since measures were never read.

Coordinates are given measures. On a coordinate grid, lengths and slopes are computed from coordinates the problem states, so they are given, not read.

Measuring is a different task. When a problem asks you to measure a drawing with a ruler or protractor, the drawing is the object. The rule governs proofs and “find” problems, where the figure is a sketch.

Where this goes next

The set-up is the first two lines of every proof on this site. Two-column and paragraph proofs fill the space between them with a reason for every step, with the Given as the first statement and “given” as its reason, and the marks on the diagram become statements in those first lines. The plans in Examples 9 and 11 needed auxiliary lines, and planning a congruence proof turns a plan like the one in Example 12 into a route. The facts behind the marks are congruent segments, congruent angles, perpendicular lines and parallel lines.

Why a proof about one named figure is a proof about all of them is the subject of deductive reasoning. The rule about the diagram is one case of a wider one: a drawing is one example, and one example proves nothing about all cases, which is why a pattern is not a proof. A theorem and its converse shows how a fact that is proved can be used later in place of a mark.

Exercises

Given and Prove

Write the Given and the Prove, using the letters suggested.

If two angles form a linear pair, then they are supplementary. Use \angle 1 and \angle 2.
If M is the midpoint of \overline{AB}, then AM = \tfrac{1}{2}AB.
Vertical angles are congruent. Use lines \overleftrightarrow{AB} and \overleftrightarrow{CD} crossing at E.
An equilateral triangle has three congruent angles. Use \triangle PQR.
If a ray bisects an angle, each of the two angles it makes is half the original. Use \overrightarrow{BD} and \angle ABC.
Two angles complementary to the same angle are congruent. Use \angle 1, \angle 2 and \angle 3, with \angle 2 the shared angle.

Reading a figure

Say whether each statement may be read from the left panel of Figure 2, follows from what may be read, or neither.

B, D and C are collinear.
D is the midpoint of \overline{BC}.
\overrightarrow{AD} lies inside \angle BAC.
m\angle ADB = 90^\circ.
BD < BC.
m\angle BAD + m\angle DAC = m\angle BAC.

Diagrams

List the marks the diagram for Exercise 4 should carry, and the marks it must not carry.
For the set-up in Example 9, a diagram marks \angle B and \angle C with matching arcs. What is wrong with it?
To set up “the diagonals of a rectangle are congruent”, someone draws a square. Name two facts the drawing shows that the Given does not contain.

Marks and words

Triangle PQR has one tick on \overline{PQ} and one on \overline{PR}, and a point M on \overline{QR} with a small square where \overline{PM} meets \overline{QR}. Write in symbols the two facts the marks give.
A figure shows a circle, a point O drawn near its middle, points X and Y on the circle, and \overline{XY} drawn through O. Nothing is marked. Say whether each may be read, must be given, or follows once something else is given: (a) X lies on the circle; (b) O is the center of the circle; (c) \overline{XY} is a diameter.
Two lines on a page run off both edges without meeting. Explain why that does not show they are parallel.

Finding what can be found

In Figure 5 with m\angle ABD = 41^\circ instead, find m\angle DBC.
In Figure 5 with m\angle ABD = 41^\circ and \angle DBE marked as a right angle, find m\angle EBC.
A figure shows A, B and C on a line with B between them, AB = 2x + 3, BC = x - 1 and AC = 20. Find x, AB and BC.
The same figure also has matching ticks on \overline{AB} and \overline{BC}, with AB = 5y - 4 and BC = 3y + 6 (and no length given for AC). Find y and AC.
Two lines cross. Two neighboring angles at the crossing measure (2x + 30)^\circ and (3x - 5)^\circ, and the first looks like a right angle. Find both angles.
In a figure like Figure 6 with \ell \parallel m marked, the two marked angles measure (3x + 5)^\circ and (4x - 15)^\circ. Find x and the angle.

Using a set-up

Each exercise uses a set-up from this lesson, with its Prove taken as proved.

In Example 8, m\angle A = (3x + 4)^\circ and m\angle B = (2x + 11)^\circ. Find x and both angles.
In Figure 1, BC = 4k - 2 and DE = k + 5. Find k, BC and DE.
In Example 10, CA = 4y - 3 and CB = 2y + 9. Find y and CA.
In Example 12, AE = 3t + 2 and EC = 5t - 8. Find t, AE and AC.
In Example 9, m\angle B = (4n - 10)^\circ and m\angle C = (2n + 30)^\circ. Find n and all three angles of the triangle.

Statements with no letters

Choose letters, write the Given and the Prove, and say what the diagram marks.

If two parallel lines are cut by a transversal, then alternate interior angles are congruent.
If two angles of one triangle are congruent to two angles of another triangle, then the third angles are congruent.
The median to the base of an isosceles triangle is perpendicular to the base. (A median joins a vertex to the midpoint of the opposite side.)

Reasoning

Explain why the diagram must never mark what the Prove says.
A proof works with \triangle ABC throughout. Explain why it proves the theorem for every triangle that fits the Given, and what could stop it doing so.
Write the converse of the statement in Exercise 1 as a set-up, and show by an example that it is false.
A set-up for the theorem in Example 9 reads: Given. \triangle ABC with \overline{AB} \cong \overline{AC}; \angle B \cong \angle C. Prove. \triangle ABC is isosceles. Say what is wrong with it.
Explain why the rule allows reading that D is between B and C, but not that D is the midpoint of \overline{BC}.
Using the left panel of Figure 2, prove that m\angle ADB + m\angle ADC = 180^\circ. Say for each step whether it is read from the figure or cited.
Explain why a figure can suggest that \overline{AD} \perp \overline{BC} but can never show it, even drawn with perfect instruments.

Answers

Given. \angle 1 and \angle 2 form a linear pair. Prove. \angle 1 and \angle 2 are supplementary, that is, m\angle 1 + m\angle 2 = 180^\circ.
Given. M is the midpoint of \overline{AB}. Prove. AM = \tfrac{1}{2}AB.
As an if-then: if two lines cross, the angles opposite each other at the crossing are congruent. Given. \overleftrightarrow{AB} and \overleftrightarrow{CD} meet at E, with E between A and B and between C and D. Prove. \angle AEC \cong \angle BED and \angle AED \cong \angle BEC. Both pairs, since the words cover every vertical pair.
If a triangle is equilateral, its three angles are congruent. Given. \triangle PQR with PQ = QR = RP. Prove. \angle P \cong \angle Q \cong \angle R.
Given. \overrightarrow{BD} bisects \angle ABC. Prove. m\angle ABD = \tfrac{1}{2}m\angle ABC and m\angle DBC = \tfrac{1}{2}m\angle ABC.
Given. \angle 1 and \angle 2 are complementary; \angle 3 and \angle 2 are complementary. Prove. \angle 1 \cong \angle 3.
Read. Collinearity is a position fact.
Neither. Midpoint means BD = DC, a measure fact, and there are no ticks.
Read. Whether a ray is inside an angle is a position fact.
Neither. It needs a square at D.
Follows. D is between B and C, so BC = BD + DC, and DC > 0.
Follows. \overrightarrow{AD} is inside \angle BAC (read), so the angle addition postulate gives the sum.
One tick on each of the three sides, the same number on all three, since the Given makes all three congruent. No arcs at P, Q or R: congruent angles are the Prove.
Matching arcs say \angle B \cong \angle C, which is the Prove. The diagram then assumes the conclusion, and a proof reading it would be circular.
Any two of: all four sides congruent; diagonals perpendicular; diagonals bisecting the corner angles. A rectangle need have none of these, so draw one clearly longer than it is wide.
PQ = PR (that is, \overline{PQ} \cong \overline{PR}), and \overline{PM} \perp \overline{QR}.
(a) is read: it is incidence. (b) must be given, in words or by calling the circle “circle O”, because a center is equidistant from every point of the circle, a measure fact. (c) follows once (b) is given: a diameter is a chord through the center, and that X, O and Y are collinear with O between X and Y is read. Without (b), (c) cannot be known.
Parallel lines never meet anywhere. The page shows only a piece of each, and lines that meet far off the page look exactly like parallel lines on it.
139^\circ. \angle ABD and \angle DBC form a linear pair: 180^\circ - 41^\circ = 139^\circ.
49^\circ. m\angle DBE + m\angle EBC = 139^\circ, so m\angle EBC = 139^\circ - 90^\circ = 49^\circ.
x = 6, AB = 15, BC = 5. B is between, so (2x + 3) + (x - 1) = 20, 3x + 2 = 20. Check: 15 + 5 = 20.
y = 5 and AC = 42. The ticks give 5y - 4 = 3y + 6, so 2y = 10. Each piece is 21, and AC = 21 + 21 = 42.
92^\circ and 88^\circ. Neighbors form a linear pair: (2x + 30) + (3x - 5) = 180, so 5x = 155 and x = 31. The one that looked right is 2(31) + 30 = 92^\circ.
x = 20, and each angle is 65^\circ. Corresponding angles are equal: 3x + 5 = 4x - 15.
x = 15, m\angle A = 49^\circ and m\angle B = 41^\circ. The Prove makes them complementary: (3x + 4) + (2x + 11) = 90, so 5x = 75.
k = 6, BC = 22 and DE = 11. The Prove gives DE = \tfrac{1}{2}BC, so k + 5 = 2k - 1. Check: 22 = 2 \times 11.
y = 6 and CA = 21. The Prove gives 4y - 3 = 2y + 9, so 2y = 12; CB = 2(6) + 9 = 21 too.
t = 5, AE = 17 and AC = 34. The Prove gives 3t + 2 = 5t - 8, so 2t = 10. Then EC = 17 and AC = AE + EC = 34.
n = 20, m\angle B = m\angle C = 70^\circ and m\angle A = 40^\circ. The Prove gives 4n - 10 = 2n + 30, so 2n = 40; the angle sum gives 180^\circ - 140^\circ = 40^\circ.
Given. \ell \parallel m; line t crosses \ell at P and m at Q; A on \ell and B on m lie on opposite sides of t. Prove. \angle APQ \cong \angle BQP. The diagram marks \ell and m with matching arrowheads and puts no arcs at P or Q. The other alternate pair follows from this one by linear pairs.
Given. \triangle ABC and \triangle DEF; \angle A \cong \angle D; \angle B \cong \angle E. Prove. \angle C \cong \angle F. Two separate triangles of different sizes, since nothing says they match; one arc at A and D, two at B and E, nothing at C and F.
Given. \triangle ABC with \overline{AB} \cong \overline{AC}; M is the midpoint of \overline{BC}. Prove. \overline{AM} \perp \overline{BC}. The base is \overline{BC}, the side between the ends of the congruent sides. One tick on each of \overline{AB} and \overline{AC}, two on each of \overline{BM} and \overline{MC}, and no square at M.
A mark is part of the Given. Marking the Prove puts the conclusion among the assumptions, and a proof could then quote it in one line and have shown nothing.
The letters name one triangle, but a proof that uses only the Given uses nothing true of that triangle alone, so the same steps apply word for word to any triangle the Given fits. It fails if a step leans on something true only of ABC as drawn: a measure read off the picture, or a fact that crept into the Given.
Given. \angle 1 and \angle 2 are supplementary. Prove. \angle 1 and \angle 2 form a linear pair. False: an angle of 100^\circ in one corner of a page and an angle of 80^\circ in another are supplementary but share no vertex, so they are not a linear pair.
Its Given contains \angle B \cong \angle C, the theorem’s conclusion, and its Prove is the theorem’s hypothesis, since \overline{AB} \cong \overline{AC} is what isosceles means. The Prove is already in the Given, so a proof of it shows nothing. The right set-up is Given. \triangle ABC with \overline{AB} \cong \overline{AC}. Prove. \angle B \cong \angle C.
Betweenness is a position fact: every figure fitting “D on \overline{BC}” has D between B and C. Midpoint is a measure fact, BD = DC, true in some of those figures and false in others (Figure 3), so it must be given.
B, D and C are collinear with D between B and C (read). So \overrightarrow{DB} and \overrightarrow{DC} are opposite rays, and \angle ADB and \angle ADC share side \overrightarrow{DA} (read): they form a linear pair (definition). Linear pairs are supplementary (cited).
A drawing is one figure; the claim has to hold for every figure fitting the given, and in some of those the angle is not 90^\circ. And no drawing tells 90^\circ from 89.99^\circ, so even a perfect one could not settle it.