Geometry · G.CO.D.12

Copying a segment and an angle

You have a segment, or an angle, drawn on paper, and no ruler marks and no protractor. How do you make an exact copy of it somewhere else?

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Segment length, addition and congruenceMath 8 Congruent angles, angle addition and bisectorsMath 7

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Bisecting a segment and an angleGeometry Constructing a triangle from given partsGeometry

Where you are 8 in Compass and straightedge constructions

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Bisecting a segment and an angle Constructing a triangle from given parts

Copying a segment and an angle

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When you reach for it

A construction is a drawing made with exactly two tools. The straightedge draws the line through two points you already have. It carries no marks, so it cannot measure anything. The compass draws a circle, or a piece of one called an arc, about a center you choose, and it holds its opening when you lift it. Nothing else is allowed: no scale on a ruler, no protractor, no judging by eye.

The restriction is the point. A measured drawing is only as good as the ruler and the eye that read it. A construction is exact, because every step is justified by a definition or a theorem, and the justification does not depend on how steady your hand is.

Copying is the first move of all constructions. Reach for it whenever a later construction needs a length or an angle you already have, placed somewhere new: laying lengths end to end to add them, building a triangle with given parts, or making an angle equal to one somewhere else in the figure, which is how a parallel line through a point is drawn.

A copy is a congruent figure placed where you want it. From Segment length, addition and congruence, two segments are congruent when they have the same length; from Congruent angles, angle addition and bisectors, two angles are congruent when they have the same measure. So copying a segment means producing a segment of the same length, and copying an angle means producing an angle of the same measure, without ever finding out what that length or measure is.

The procedure

Copying a segment

You are given \overline{AB} and a ray starting at a point P. The goal is a point Q on the ray with \overline{PQ} \cong \overline{AB}.

Set the compass to AB. Put the compass point on A and open it until the pencil sits on B. The opening is now the distance AB. Nothing has been measured; the compass simply holds the distance.

Swing an arc from P. Without changing the opening, move the point to P and draw an arc that crosses the ray. Call the crossing Q.

Why PQ = AB. The arc is part of the circle with center P and radius AB, and a circle is exactly the set of points at that one distance from its center. Q is on the arc, so PQ = AB, and segments of equal length are congruent: \overline{PQ} \cong \overline{AB}.

Figure 1The compass is set to AB and swung from P. Every point of the arc is AB away from P, so PQ = AB.

The ray starts at the center of the circle, so it leaves the circle at exactly one point. Q is determined, and there is no choice to make. On a whole line through P the circle is met twice, once on each side of P, and either point gives a copy.

The same move, made again from Q, adds lengths. Copy a second segment from Q, taking the crossing on the side away from P, and its far end R has Q between P and R, so PR = PQ + QR by segment addition. Take the crossing on the side toward P instead and the lengths subtract.

Copying an angle

You are given \angle ABC, with vertex B, and a ray \overrightarrow{PQ}. The goal is a ray from P that makes an angle with \overrightarrow{PQ} congruent to \angle ABC.

The compass cannot hold an angle, only a distance. So the angle is turned into a triangle, which is fixed by its three sides; the three sides are copied as distances, and the angle comes with them.

An arc at the old vertex. With the point on B and any convenient opening r, draw an arc crossing both arms: it meets \overrightarrow{BA} at D and \overrightarrow{BC} at E. Both points lie on the circle with center B and radius r, so BD = BE = r.

The same arc at the new vertex. Keep the opening r. Put the point on P and draw a long arc that crosses \overrightarrow{PQ} at F and sweeps round into the side where the new angle should open. Now PF = r, and every point of this arc is at distance r from P.

Carry the width of the angle. Set the compass to DE: point on E, pencil on D. This is the distance that tells a wide angle from a narrow one, since two angles whose arms are cut off at the same length r differ only in how far apart the cut points are. With that opening, put the point on F and draw an arc crossing the long arc. Call the crossing G. It is at distance r from P and at distance DE from F.

Draw the ray. The straightedge draws \overrightarrow{PG}, and \angle GPF \cong \angle ABC.

Figure 2Copying \angle ABC at P. The triangles \triangle DBE and \triangle GPF, each closed by a dashed segment, have three pairs of equal sides, so their angles at B and P are equal.

Why it works. Join D to E and F to G (the dashed segments in the figure). The triangles \triangle DBE and \triangle GPF have

BD = PG = r, \quad BE = PF = r, \quad DE = GF

so \triangle DBE \cong \triangle GPF by SSS, with D matched to G, B to P and E to F. Corresponding parts of Congruent triangles are congruent, so \angle DBE \cong \angle GPF. Finally \angle DBE is the same angle as \angle ABC, because D lies on the arm \overrightarrow{BA} and E on the arm \overrightarrow{BC}, and naming an angle by different points of the same arms does not change it. Since F is on \overrightarrow{PQ}, \angle GPF is also \angle GPQ.

The same fact can be seen with rigid motions. A slide and a turn carry B to P and the arm \overrightarrow{BC} onto \overrightarrow{PQ}, so E lands on F. Rigid motions keep distances, so D lands on a point that is r from P and DE from F: one of the points where the two arcs cross. If it lands on the wrong side, a reflection in the line PQ moves it to the other. Rigid motions keep angle measure, so the image of \angle ABC is the angle at P that was drawn.

Two answers, one on each side. The two circles, center P with radius r and center F with radius DE, cross at two points, mirror images of each other in the line PQ. Each gives a ray making an angle congruent to \angle ABC with \overrightarrow{PQ}, one on each side of the line. Drawing the long arc only on the side you want makes the choice.

The opening r is yours to choose, and any choice works, because the argument never used its size: only that the same r is used at both vertices. A larger r is simply easier to draw accurately.

Worked examples

Example 1

Laying two lengths end to end

\overline{AB} is 3.2 cm long and \overline{CD} is 2.5 cm long. Construct a segment whose length is AB + CD, and say how long it is.

Solution

Draw a ray from a point P, long enough for both lengths. Set the compass to AB and swing an arc from P across the ray; call the crossing Q. By the definition of a circle, PQ = AB = 3.2 cm.

Now set the compass to CD and put its point on Q. The circle about Q meets the line through P and Q twice, once on each side of Q. Take the crossing R on the side away from P. Then QR = CD = 2.5 cm, and Q lies between P and R, so segment addition gives

PR = PQ + QR = 3.2 + 2.5 = 5.7 \text{ cm}

The numbers are only there to check the result. The construction never read a scale, and it produces a segment of length AB + CD whatever the two lengths are.

Example 2

A multiple and a difference

With the same segments, AB = 3.2 cm and CD = 2.5 cm, construct a segment of length 3 \cdot CD - AB, and find its length.

Solution

Set the compass to CD once and leave it set. From P mark Q_1 on a ray; from Q_1 mark Q_2, further along; from Q_2 mark Q_3, further still. Each step is a copy of \overline{CD}, and each new point lies beyond the one before, so segment addition, used twice, gives

PQ_3 = CD + CD + CD = 3 \cdot CD = 7.5 \text{ cm}

Now set the compass to AB, put its point on Q_3, and take the crossing R on the side back toward P, so RQ_3 = AB = 3.2 cm. Since 3.2 < 7.5, this crossing lands between P and Q_3. With R between them, PR + RQ_3 = PQ_3, and

PR = PQ_3 - RQ_3 = 7.5 - 3.2 = 4.3 \text{ cm}

Had AB been longer than 3 \cdot CD, the crossing would have fallen beyond P, on the far side from Q_3, and no segment of length 3 \cdot CD - AB would exist: that number would be negative.

Example 3

Copying an angle to a chosen side

Points P and Q lie on a line, and a point T is off the line, with m\angle TPQ = 70^\circ. Copy an angle \angle ABC of measure 48^\circ at P, with one arm along \overrightarrow{PQ}, so that the new angle opens on the side of line PQ away from T. What is m\angle TPG, where \overrightarrow{PG} is the new arm? What would it have been on the other side?

Solution

With any opening r, draw an arc about B crossing the arms at D and E. With the same r, draw an arc about P that crosses \overrightarrow{PQ} at F and sweeps into the half of the page that does not contain T. Set the compass to DE and swing an arc from F; where it crosses the first arc on the side away from T, call the point G. Draw \overrightarrow{PG}.

By the argument of the procedure, \triangle DBE \cong \triangle GPF by SSS, so \angle GPQ \cong \angle ABC and m\angle GPQ = 48^\circ.

Now \overrightarrow{PT} and \overrightarrow{PG} lie on opposite sides of line PQ, and the two angles together measure 70^\circ + 48^\circ = 118^\circ, less than 180^\circ. So \overrightarrow{PQ} runs between them, inside \angle TPG, and by the angle addition postulate

\begin{aligned} m\angle TPG &= m\angle TPQ + m\angle QPG \\ &= 70^\circ + 48^\circ = 118^\circ \end{aligned}

The other crossing, G', lies on the same side as T. Then both \angle TPQ and \angle G'PQ open from \overrightarrow{PQ} into the same side, and the smaller one, 48^\circ, sits inside the larger, so m\angle TPG' = 70^\circ - 48^\circ = 22^\circ. Both copies are correct copies of \angle ABC; only the side was chosen.

Example 4

Adding and subtracting two angles

m\angle 1 = 35^\circ and m\angle 2 = 50^\circ. Construct an angle whose measure is m\angle 1 + m\angle 2, and one whose measure is m\angle 2 - m\angle 1.

Solution

For the sum, copy \angle 1 at P with one arm along \overrightarrow{PQ}; call the new arm \overrightarrow{PR}. Then copy \angle 2 at P with one arm along \overrightarrow{PR}, choosing the side of line PR away from Q; call its new arm \overrightarrow{PS}. The two angles share the arm \overrightarrow{PR} and do not overlap, and their total, 85^\circ, is less than 180^\circ, so R lies inside \angle QPS. The angle addition postulate gives

\begin{aligned} m\angle QPS &= m\angle QPR + m\angle RPS \\ &= 35^\circ + 50^\circ = 85^\circ \end{aligned}

Figure 3Two copied angles placed side by side, sharing the ray \overrightarrow{PR}. Together they make \angle QPS.

For the difference, copy both angles at P with one arm along \overrightarrow{PQ} and both on the same side of line PQ: \angle 1 gives \overrightarrow{PR} and \angle 2 gives \overrightarrow{PS'}. The smaller angle sits inside the larger, so \overrightarrow{PR} lies inside \angle QPS', and

\begin{aligned} m\angle RPS' &= m\angle QPS' - m\angle QPR \\ &= 50^\circ - 35^\circ = 15^\circ \end{aligned}

Example 5

Copying a whole triangle

Given \triangle ABC, construct a triangle congruent to it by copying only three of its parts.

Solution

Draw a ray from a point D. Copy \overline{AB} onto it, giving E with DE = AB. Copy \angle BAC at D with one arm along \overrightarrow{DE}, on either side. Along the new arm copy \overline{AC}, giving F with DF = AC. Finally the straightedge joins E to F.

The two triangles have \overline{DE} \cong \overline{AB}, \angle EDF \cong \angle BAC and \overline{DF} \cong \overline{AC}: two sides and the angle between them. By SAS, \triangle DEF \cong \triangle ABC, with A matched to D, B to E and C to F. So \overline{EF} \cong \overline{BC}, \angle E \cong \angle B and \angle F \cong \angle C, even though none of these three was copied. A congruence criterion is exactly a list of parts whose copies force the rest.

Example 6

The third angle of a triangle, without the triangle

Two angles of a triangle measure 48^\circ and 69^\circ. Construct the third angle without drawing the triangle, and find its measure.

Solution

Draw a line and mark P on it, with Q on the line to one side of P and T to the other, so \overrightarrow{PQ} and \overrightarrow{PT} are opposite rays. Copy the 48^\circ angle at P with one arm along \overrightarrow{PQ}, giving \overrightarrow{PR}. Copy the 69^\circ angle at P with one arm along \overrightarrow{PR}, on the side away from Q, giving \overrightarrow{PS}. As in Example 4, m\angle QPS = 48^\circ + 69^\circ = 117^\circ.

Figure 4Two angles of a triangle copied side by side on a line. What is left of the straight angle is the third angle.

The angles \angle QPS and \angle SPT share the arm \overrightarrow{PS}, and their other arms are opposite rays, so they form a linear pair and are supplementary:

m\angle SPT = 180^\circ - 117^\circ = 63^\circ

By the Triangle angle sum, the third angle of the triangle is 180^\circ - 48^\circ - 69^\circ = 63^\circ as well, so \angle SPT is congruent to it. The three angles of a triangle, copied side by side, fill a straight angle exactly, which is a picture of the angle sum theorem itself.

Boundaries

Only the two tools, and only for drawing. The lengths and measures in the examples are there to check the results; no step reads them. A ruler’s scale or a protractor would make each copy a measurement, and a measurement is only approximate.

The compass must hold its opening between the two places. Every copy rests on the radius set at one place being the radius used at the other. Euclid’s Elements does not take even this for granted: Book I, Proposition 2 places at a given point a segment equal to a given one, drawing only circles about known centers through known points. So a compass that closed whenever it was lifted would still be enough; a compass that holds its setting does the same job in one step.

A ray gives one copy, a line gives two. From the end of a ray the arc crosses it once. On a whole line it crosses on both sides of P, and you choose.

The first arc must cross both arms. Any opening r > 0 does, since arms are rays and go on as far as needed; a short drawn arm is simply extended with the straightedge first.

The same opening at both vertices. If the arc at P has a different radius from the arc at B, the triangles \triangle DBE and \triangle GPF do not have matching sides, SSS says nothing, and the angle drawn is wrong.

The angle must lie strictly between 0^\circ and 180^\circ. For a straight angle, D, B and E lie on one line and there is no triangle; the copy is just the ray opposite \overrightarrow{PQ}. Every other angle gives DE < 2r, by the Triangle inequality in \triangle DBE, and that is what makes the two arcs cross at two points.

An angle copy has two positions. One on each side of the line it is copied onto, mirror images of each other. Say which side, or both answers are right.

Sums and differences have limits. A difference of lengths, AB - CD, exists only when CD < AB. Two angles placed side by side make a single angle only when their total is less than 180^\circ. At exactly 180^\circ the two outer arms are opposite rays, and above it the angle between the outer arms, measured the short way, is 360^\circ minus the total.

Where this goes next

Almost every other construction is made of these two copies. Drawing two arcs of the same radius from the two ends of a segment is the start of bisecting a segment and an angle, and copying one side and then swinging arcs of the other two lengths is constructing a triangle from given parts. Copying an angle so that a new line makes the same angle with a transversal as an old one gives a parallel line through a point, justified by the converse of the corresponding angles postulate. Marking the same length again and again along a ray, as in Example 2, is the first half of dividing a segment into equal parts.

The way each step here was defended, by a circle’s definition or by a pair of congruent triangles, is the whole method of justifying a construction, and the same equal circles, drawn through each other’s centers, give the equilateral triangle, square and hexagon.

Exercises

Copying segments

AB = 4.1 cm and CD = 2.7 cm. Copy \overline{AB} onto a ray from P, ending at Q, then copy \overline{CD} from Q on the side away from P, ending at R. How long is \overline{PR}?
AB = 4.1 cm and CD = 2.7 cm. Copy \overline{AB} onto a ray from P, ending at Q, then copy \overline{CD} from Q on the side back toward P, ending at R. How long is \overline{PR}?
AB = 4.1 cm and CD = 2.7 cm. Describe how to construct a segment of length 2 \cdot AB + CD, and give its length.
A whole circle is drawn with center P and the compass set to a segment 6 cm long. A line through P crosses the circle at two points. How far apart are they?
On a coordinate grid, A = (1, 1) and B = (4, 5). The compass is set to AB and its point placed at P = (2, 0). (a) Where does the arc cross the ray from P pointing straight up, in the positive y direction? (b) Where does the whole circle cross the x-axis?

Copying angles

m\angle 1 = 38^\circ and m\angle 2 = 57^\circ. \angle 1 is copied at P onto \overrightarrow{PQ}, giving \overrightarrow{PR}, and \angle 2 is copied at P onto \overrightarrow{PR} on the side away from Q, giving \overrightarrow{PS}. Find m\angle QPS.
m\angle 1 = 38^\circ and m\angle 2 = 57^\circ. Both are copied at P onto \overrightarrow{PQ}, on the same side of line PQ, giving \overrightarrow{PR} and \overrightarrow{PS} respectively. Find m\angle RPS.
An angle of 64^\circ is copied three times around a point P, starting from \overrightarrow{PQ}, each copy sharing an arm with the one before and turning the same way. What angle does the last ray make with \overrightarrow{PQ}?
An angle of 72^\circ is copied five times around a point P in the same way, starting from \overrightarrow{PQ}. Where does the last ray land, and why?
Two angles of a triangle measure 41^\circ and 86^\circ. They are copied side by side at a point of a line, as in Example 6. What angle is left over on the line?
An angle of 112^\circ is copied at P with one arm along \overrightarrow{PQ}, and \overrightarrow{PT} is the ray opposite \overrightarrow{PQ}. What angle does the new arm make with \overrightarrow{PT}?
P and Q lie on a line, T is off it, and m\angle TPQ = 55^\circ. An angle of 40^\circ is copied at P with one arm along \overrightarrow{PQ}, giving a new arm \overrightarrow{PG}. Find m\angle TPG (a) when G is on the opposite side of line PQ from T, (b) when G is on the same side as T.

Triangles from copies

Kofi copies \overline{AB} of \triangle ABC onto a ray as \overline{DE}, copies \angle A at D and \angle B at E on the same side of \overline{DE}, and lets the two new arms meet at F. Which congruence criterion makes \triangle DEF \cong \triangle ABC? If m\angle A = 35^\circ and m\angle B = 75^\circ, find m\angle F.
Ines tries to copy \triangle ABC by copying \overline{AB}, then \angle B at the end of her copy, then swinging an arc of radius AC from the copy of A to meet the new arm. Explain why this may not produce a copy of \triangle ABC.

Reading the construction

In copying \angle ABC, Femi sets the compass to 4 cm for the arc about B, and the points D and E turn out to be 3 cm apart. Give PF, PG and FG in his copy.
Femi copies the same angle again, but this time opens the compass to 5 cm for the arc about P, while still using 3 cm from F. Is the new angle congruent to \angle ABC? Explain.

Reasoning

Explain why the compass must not be adjusted between setting it on \overline{AB} and swinging the arc from P.
Prove that the angle-copying construction produces an angle congruent to the given one.
In copying an angle, explain why the arc from F crosses the arc about P at two points, and how the two rays they give compare.
Use a rigid motion to explain why copying a segment onto a ray works.

Answers

6.8 cm. Q lies between P and R, so PR = 4.1 + 2.7 = 6.8.
1.4 cm. Since 2.7 < 4.1, R lands between P and Q, so PR = 4.1 - 2.7 = 1.4.
10.9 cm. Leave the compass set to AB and mark it twice from P along a ray, then set it to CD and mark once more, always on the side away from P: 4.1 + 4.1 + 2.7 = 10.9.
12 cm. Both points are 6 cm from P, on opposite sides of it, so P lies between them and the distance is 6 + 6 = 12.
(a) (2, 5). From A to B is 3 across and 4 up, so by the The Pythagorean theorem \displaystyle AB = \sqrt{3^2 + 4^2} = 5, and the crossing is 5 above P. (b) (-3, 0) and (7, 0), 5 to the left and 5 to the right of P.
95^\circ. The angles share \overrightarrow{PR} and their total is under 180^\circ, so angle addition gives 38^\circ + 57^\circ = 95^\circ.
19^\circ. The smaller copy sits inside the larger, so 57^\circ - 38^\circ = 19^\circ.
168^\circ. Turning one way the last ray is 3 \cdot 64^\circ = 192^\circ round from \overrightarrow{PQ}, past the opposite ray; measured the short way the angle is 360^\circ - 192^\circ = 168^\circ.
Exactly on \overrightarrow{PQ}. Five copies turn 5 \cdot 72^\circ = 360^\circ, one full turn. The five rays split the turn into five congruent angles, the central angles of a regular pentagon.
53^\circ. Side by side the two angles fill 41^\circ + 86^\circ = 127^\circ, and the rest of the straight angle is 180^\circ - 127^\circ = 53^\circ, the triangle’s third angle.
68^\circ. The copied angle and the angle beside it form a linear pair, so 180^\circ - 112^\circ = 68^\circ.
(a) 95^\circ: \overrightarrow{PQ} lies inside \angle TPG, so 55^\circ + 40^\circ = 95^\circ. (b) 15^\circ: the 40^\circ angle sits inside the 55^\circ one, so 55^\circ - 40^\circ = 15^\circ.
ASA, and m\angle F = 70^\circ. Two angles and the side between them are copied, so ASA gives the congruence; the third angle is 180^\circ - 35^\circ - 75^\circ = 70^\circ.
She has copied two sides and an angle that is not between them, and SSA is not a congruence criterion. When AC is shorter than AB, the arc of radius AC about the copy of A can cross the new arm at two points, giving two different triangles with the same three copied parts, and only one of them need be congruent to \triangle ABC. The construction does not say which. Copying \angle A instead, the angle between the two copied sides, makes it SAS and settles it.
PF = 4 cm, PG = 4 cm, FG = 3 cm. F and G both lie on the arc about P drawn with the same opening as the arc about B, and G is on the arc about F with opening DE.
No; it comes out smaller. His triangles now have sides 4, 4, 3 and 5, 5, 3, so they are not congruent and SSS proves nothing. With the correct angle and arms cut at 5 cm, the cut points would be \displaystyle 3 \cdot \frac{5}{4} = 3.75 cm apart, the 4, 4, 3 triangle scaled up; joining them at only 3 cm pulls the arms closer together, so the angle is narrower.
The point Q is only known to be AB from P because it lies on a circle of radius AB about P. If the opening changes, the circle about P has some other radius, and PQ equals that radius rather than AB.
Let the arc of radius r about B meet the arms at D and E, the arc of radius r about P meet \overrightarrow{PQ} at F, and the arc of radius DE about F meet it at G. Then BD = PG = r, BE = PF = r and DE = GF, so \triangle DBE \cong \triangle GPF by SSS. Corresponding parts of congruent triangles are congruent, so \angle DBE \cong \angle GPF, which are \angle ABC and \angle GPQ named by other points on the same arms.
The circles have centers P and F, a distance r apart, and radii r and DE. Two circles meet at two points when the distance between centers lies strictly between the difference and the sum of the radii: |r - DE| < r < r + DE. The right-hand inequality holds because DE > 0, and the left-hand one because 0 < DE < 2r, by the triangle inequality in \triangle DBE. The two crossings are reflections of each other in the line PF, so the two rays make congruent angles with \overrightarrow{PQ}, one on each side of the line.
Slide A to P, then turn about P until the image of \overrightarrow{AB} lies along the ray. Rigid motions keep lengths, so B lands on the ray at distance AB from P. The ray holds exactly one point at that distance, the point Q where the arc crosses it, so the image of \overline{AB} is \overline{PQ}, and the two are congruent.