Proving parts congruent after the triangles

When you reach for it

Look at Figure 1. E is the midpoint of \overline{AD} and of \overline{BC}, and you are asked to prove \overline{AB} \parallel \overline{DC}. Nothing in the Given mentions either segment, and no definition turns “midpoint” into a fact about angles at A or D.

But the angles at A and D that would make the segments parallel are angles of \triangle AEB and \triangle DEC, and those two triangles can be proved congruent from what you do know. Once they are, every pair of their corresponding parts is congruent, including the pair you wanted. That is the move: prove the triangles congruent first, then claim the part, and then, often, one more step.

Reach for it whenever the Prove names a side or an angle, or something built from one (a midpoint, a bisector, parallel lines, perpendicular lines), and the Given says nothing about it directly. Almost nobody wants congruent triangles for their own sake. A navigator wants the distance to a ship; a proof wants two segments equal, or two angles. The triangles are a tool for reaching a part that could not be measured or was not given.

Writing the first half is proving triangles congruent, and choosing which triangles to use is planning a congruence proof. This lesson is about the second half and how far it can carry you: the result that lets a part out of a congruence, the procedure that uses it, and the two stage proof, where the part won from one pair of triangles is what proves a second pair congruent.

The procedure

Corresponding parts of congruent triangles

The congruence criteria prove two triangles congruent from three matching pairs of parts. So a proof about triangles usually runs in two moves. A criterion gets you in: three known pairs give the congruence. The result in this section gets you out: the congruence gives every pair, including the three nobody checked.

Definition. Two triangles are congruent when their vertices can be matched so that all six pairs of parts are congruent (congruence statements and correspondence). Parts that are matched with each other are corresponding parts, and the statement \triangle ABC \cong \triangle DEF names the matching by letter order: A with D, B with E, C with F.

The result. Corresponding parts of congruent triangles are congruent. If \triangle ABC \cong \triangle DEF, then

and so, in numbers, AB = DE, m\angle A = m\angle D, and so on for all six pairs.

Many books shorten the sentence to its initials, CPCTC. In a two-column proof this lesson writes the reason as “corresponding parts of congruent triangles”, followed by the line that holds the congruence statement.

The order is fixed. The congruence statement must come first, justified by a criterion that uses only given or already proved parts. A part may be claimed as corresponding only after that line, and only for the part that the letter order pairs it with.

The picture. A congruence criterion guarantees that one triangle is an exact copy of the other. The three parts it checks are only the evidence; an exact copy agrees with the original in every measurement.

In Figure 2 the ink marks are what SAS was given: two sides and the angle between them. The gray marks are the other three pairs. Nobody measured them; they are congruent because the triangles are.

Why it is true. There is nothing to prove beyond the definition. “Congruent triangles” was defined to mean triangles whose six pairs of corresponding parts are congruent, and every definition works in both directions (definitions are biconditionals): six matching pairs make the triangles congruent, and congruent triangles have six matching pairs. The result is the second reading.

The criteria make it useful. SSS, SAS, ASA and AAS each conclude a congruence statement from three pairs, and a congruence statement carries all six, so the three unchecked pairs were part of what the criterion concluded. The order matters because a reason may only point back to earlier lines (justifying each step); a part cited before its congruence leans on nothing.

Claiming a part, step by step

Find two triangles that hold the part. The segment or angle you want must be a side or an angle of one triangle, and the thing you want it congruent to must be the matching part of the other. If no such pair of triangles is in the figure, this move is not yet available.

Prove the triangles congruent from other parts. Use SSS, SAS, ASA or AAS, as in the congruence criteria, with three pairs you can actually establish. The pair you are after can never be one of the three: you do not have it yet, and using it would make the proof lean on its own conclusion. The order is fixed: criterion first, part second.

Write the congruence statement so the letters line up. The statement \triangle AEB \cong \triangle DEC pairs A with D, E with E and B with C, and every part is read off from that pairing, as in congruence statements and correspondence. Before you claim a part, check that it really corresponds: here \overline{AB} goes with \overline{DC}, but \overline{AE} goes with \overline{DE}, not with \overline{EC}.

Claim the part, citing the triangle line. Two triangles are congruent exactly when all six pairs of corresponding parts are congruent, so once the triangles are, each of the six pairs is: that is the result of the subsection above. In a two-column proof the reason reads “corresponding parts of congruent triangles: line 6”, naming the line where the triangles were proved congruent.

Carry the part to the conclusion you need. Often the part is not the end. These are the common next steps, each a definition read backwards or a converse:

• A midpoint. Claim \overline{AM} \cong \overline{MB}, check that M is on \overline{AB}, and cite the definition of midpoint.
• An angle bisector. Claim \angle ABD \cong \angle DBC, check that D is inside \angle ABC, and cite the definition of angle bisector.
• Parallel lines. Claim a pair of alternate interior angles along a transversal, and cite the converse of the Alternate Interior Angles Theorem.
• Perpendicular lines. Claim that the two angles of a linear pair are congruent, and cite the result the lesson on perpendicular lines calls equal neighbors: if two lines cross and two neighboring angles are equal, the lines are perpendicular. The reason reads “equal neighbors”, followed by the congruence line and the linear pair line.

Definitions can run in this direction because every definition is a biconditional. The position facts (a point on a segment, a ray inside an angle, three points on a line) come from the Given or the diagram, never from the congruence.

One congruence gives at most three new pairs, since three of its six pairs were used to prove it. If the part you need is not among those three, you need a different pair of triangles, and that is where the two stage proof below begins.

Two stage congruence proofs

Sometimes the part you want sits in two triangles that the Given, the shared sides and the vertical angles do not prove congruent in one go. Often that fails by exactly one part. The triangles holding the Prove, call them the target pair, have two of the three parts a criterion needs, and the third is nowhere in the Given.

The missing part usually belongs to a second pair of triangles in the same figure, a helper pair, which the Given can prove congruent. Prove the helper pair first, claim the part you need because corresponding parts of congruent triangles are congruent, and hand it to the target pair. That is a two stage proof: one congruence supplies a part for another.

Start from the Prove, not the Given. Find the two triangles that contain the parts to be proved. List what they already have: given parts, a side they share (the reflexive property), vertical angles, halves from a midpoint or a bisector, angles from parallel lines. Name the criterion those parts are closest to, and say exactly which part is missing; planning a congruence proof does this for a single pair. If nothing is missing, the proof has one stage and none of this is needed.

Look for a helper pair that contains the missing part. It must be a pair of triangles in which the missing part is one of the six corresponding parts, and whose congruence the Given can reach without the target pair. In a kite, the target pair is often the small triangles at one end of a diagonal, and the helper pair is the large triangles on either side of the other diagonal.

Prove the helper pair, then claim the part. A congruence statement says all six pairs of parts match, so any one of them may be written on the next line, with the reason “corresponding parts of congruent triangles” and the line of the congruence. Read the letters of the statement to be sure the two parts really correspond.

Carry the part across. Often it is the same angle under another name: in Figure 3, E lies on \overline{AC}, so \overrightarrow{AE} is \overrightarrow{AC} and \angle BAE is \angle BAC. A diagram may supply that position fact. Sometimes one more step is needed, such as a supplement, a vertical angle or a subtraction, and that step gets its own line and reason.

Finish the target pair, then claim the Prove. The target congruence cites the helper’s part by its line number, so the helper congruence must stand higher up the page. A part of the target pair may never be claimed before the target pair is proved.

Worked examples

Boundaries

Triangles first, never the part first. Corresponding parts may be claimed only after a line proves the triangles congruent, and only by citing that line. Using the wanted part to prove the triangles congruent runs in a circle (Example 3). The same holds one level up: in a two stage proof the helper pair must come from the Given, and if its congruence needs a part of the target pair, the proof proves nothing.

Only corresponding parts. A part of one triangle is paired with exactly one part of the other, the one the congruence statement puts in its place, whatever the figure looks like. Two segments that look equal, that sit in the same place on the page, or that happen to be in the two triangles, are not thereby a corresponding pair. In Example 11, \triangle ABC \cong \triangle ADC gives \angle BAC \cong \angle DAC, because A matches A and B matches D. It does not give \angle BAC \cong \angle DCA: A is matched with A, not with C.

The congruence must be real. Known pairs of parts prove triangles congruent only through a criterion, and even three pairs work only in the arrangements the criteria allow. SSA and AAA prove nothing (why AAA and SSA fail), so they leave no parts to claim, however alike the triangles look.

The follow-on steps need their position facts. A midpoint must lie on its segment, a bisecting ray must lie inside its angle (Example 6 shows how that can fail), and a linear pair needs three points on one line. Congruent parts alone do not supply them. Likewise, an angle handed over under one name counts for a target pair only if a position fact makes it the same angle, as with E on \overline{AC} in Example 11, or a theorem makes it congruent, as with the supplements in Example 14.

The six parts, and only those. One congruence yields three sides and three angles, and three of those were spent proving it. Other segments, such as a median of each triangle, are not among the six; claiming them takes another pair of congruent triangles, which is what a two stage proof supplies.

Two stages is not a goal. If the target pair can be proved congruent from the Given directly, one stage is the better proof (Exercise 52). Some figures need three stages; each one follows the same rule, higher up the page than the stage that uses it.

Any polygon, but not the criteria. Congruent polygons of any kind also have congruent corresponding parts, by the same definition read backwards. What does not carry over is the shortcut: there is no criterion for quadrilaterals as short as SSS (Exercise 50).

Where this goes next

Nearly every proof that two segments or two angles are congruent runs through a pair of congruent triangles and ends with the reason “corresponding parts of congruent triangles”. Two things make the move harder. The triangles may overlap, sharing an angle or a stretch of side, as in proofs with overlapping triangles, where the helper and target pairs of a two stage proof often sit on top of each other. And for right triangles, HL adds one more way in.

The same move proves theorems. A single auxiliary segment splits an isosceles triangle into two congruent halves, and corresponding parts then give the base angles, in proving the Base Angles Theorem. Every point on a perpendicular bisector is equally far from the endpoints (the perpendicular bisector theorem), by the congruence of Exercise 44, and a two stage proof gives the converse, that a point equidistant from the ends of a segment lies on its perpendicular bisector.

Two stage proofs are how the properties of quadrilaterals are proved once triangles are in hand: that the diagonals of a kite are perpendicular (trapezoids and kites), that the diagonals of a rhombus are perpendicular bisectors (rectangles, rhombuses and squares), and, since a diagonal cuts a parallelogram into two congruent triangles, almost every property of parallelograms, in proofs about parallelograms.

Exercises

Reading off the parts

1. \triangle ABC \cong \triangle DEF by ASA, from \angle A \cong \angle D, \overline{AB} \cong \overline{DE} and \angle B \cong \angle E. Name the three pairs of parts that were not given.
2. In Figure 2, suppose only the gray marks had been given. Which criterion would prove the triangles congruent, and which pairs would then follow?
3. \triangle PQR \cong \triangle PSR, with side \overline{PR} shared. Which side corresponds to \overline{QR}, and which angle corresponds to \angle QPR?
4. A proof reaches \triangle JKL \cong \triangle TRS on line 6. Write line 7, claiming the partner of \overline{KL}, and line 8, claiming the partner of \angle L, each with its reason.
5. In Figure 1, \triangle AEB \cong \triangle DEC has been proved by SAS, as in Example 4. Name the three pairs of corresponding parts it gives that the midpoints and the crossing did not.

Finding measures

6. \triangle GHK \cong \triangle LMN, with m\angle G = 38^\circ, m\angle H = (4y - 6)^\circ and m\angle M = (2y + 30)^\circ. Find y and every angle of \triangle LMN.
7. \triangle PQR \cong \triangle STU, with PQ = 8, QR = 11 and SU = 13. Find the perimeter of \triangle STU.
8. \triangle ABC \cong \triangle DEF, with AB = 2a + b, BC = a - b, DE = 13 and EF = 2. Find a and b.
9. \triangle JKL \cong \triangle MNP, with JK = x^2 - 5 and MN = 4x. Find x and JK.
10. In Example 9, suppose instead that the observers, still 300 m apart, find sight lines of 66^\circ at A and 49^\circ at B, and measure AT = 250 m. Find AS, m\angle ATB and m\angle ASB.

In Exercises 11 and 12, E is the midpoint of both segments in Figure 1.

11. AB = 2x + 7 and DC = 4x - 9. Find x and AB.
12. m\angle ABE = (5y - 4)^\circ and m\angle DCE = (3y + 20)^\circ. Find y and m\angle ABE.
13. In Figure 7, with the Given of Example 8, AB = 38, PA = 3z + 5 and PB = 5z - 11. Find AM, z and PA.
14. In Figure 7, with the Given of Example 8, m\angle APB = 70^\circ. Find m\angle APM, m\angle PMA and m\angle PAM.
15. In Figure 15(b), \triangle ABC \cong \triangle ADC has been proved, m\angle BAC = (4t + 3)^\circ and m\angle DAC = (6t - 17)^\circ. Find t and m\angle BAD.

Before or after

16. Kai proves \triangle ABC \cong \triangle DEF by SAS, then writes \overline{AC} \cong \overline{EF} with the reason “corresponding parts of congruent triangles”. What is wrong, and what should the line say?
17. Two triangles have AB = DE = 7, BC = EF = 9 and m\angle B = m\angle E = 50^\circ. Mia writes AC = DF, “corresponding parts of congruent triangles”. What line must come before hers, and with what reason?

Two-column proofs

18. Use Figure 1, ignoring its ticks. Given. \overline{AB} \parallel \overline{DC}; E is the midpoint of \overline{AD}. Prove. E is the midpoint of \overline{BC}.
19. Use Figure 15(a). Given. \overline{AB} \parallel \overline{CD}; \overline{AB} \cong \overline{CD}. Prove. \overline{AD} \cong \overline{CB}.
20. Use Figure 15(a) with the same Given. Prove. \overline{AD} \parallel \overline{CB}.
21. Use Figure 15(b). Given. \overline{AB} \cong \overline{AD}; \overline{CB} \cong \overline{CD}. Prove. \overrightarrow{AC} bisects \angle BAD.
22. Use Figure 15(c). Given. M is on \overline{AB}; \overline{PM} \perp \overline{AB}; \angle PAM \cong \angle PBM. Prove. M is the midpoint of \overline{AB}.
23. Use Figure 15(a), ignoring its arrowheads. Given. \overline{AB} \cong \overline{CD}; \overline{AD} \cong \overline{CB}. Prove. \overline{AB} \parallel \overline{CD}.
24. Use Figure 15(b) for its shape only: its tick marks do not apply here, only this Given. Given. \overline{CB} \cong \overline{CD}; \overrightarrow{CA} bisects \angle BCD. Prove. \angle BAC \cong \angle DAC.
25. Use Figure 15(b) for its shape only, as in Exercise 24. Given. \overline{AC} bisects \angle BAD and \angle BCD. Prove. \overline{AB} \cong \overline{AD}.

Finishing a proof

26. Use Figure 1, ignoring its ticks. Given. \overline{AB} \parallel \overline{DC}; \overline{AB} \cong \overline{DC}. Prove. E is the midpoint of \overline{AD}. Give the reason for each line: (1) \overline{AB} \parallel \overline{DC}; (2) \angle BAE \cong \angle CDE; (3) \angle ABE \cong \angle DCE; (4) \overline{AB} \cong \overline{DC}; (5) \triangle AEB \cong \triangle DEC; (6) \overline{AE} \cong \overline{DE}; (7) E is the midpoint of \overline{AD}.
27. Use Figure 5, ignoring its marks. Given. C is the midpoint of \overline{AE} and of \overline{BD}; \overline{AB} \perp \overline{BD}. Prove. \overline{ED} \perp \overline{BD}. A proof begins (1) C is the midpoint of \overline{AE} (given); (2) \overline{AC} \cong \overline{EC} (definition of midpoint: line 1); (3) C is the midpoint of \overline{BD} (given); (4) \overline{BC} \cong \overline{DC} (definition of midpoint: line 3). Finish it.
28. Example 4 reaches \overline{AB} \parallel \overline{DC} through the transversal \overline{AD}. Keep its lines 1 to 6 and write a new line 7 and line 8 that reach the same Prove through the transversal \overline{BC}.

Two stages: finding the helper pair

For each, name the target pair, the part it is missing, and the helper pair that supplies it, with the helper’s criterion. Do not write the proof.

29. Figure 10, with the Given of Example 11. Prove. \angle CEB \cong \angle CED.
30. Figure 12 without its tick marks, where now the Given is \overline{AB} \parallel \overline{DC} and \overline{AB} \cong \overline{DC}, with \overline{AD} and \overline{BC} crossing at E and \overline{FG} through E. Prove. \overline{AF} \cong \overline{DG}.
31. Figure 13, with the Given of Example 14. Prove. \angle BED is a right angle.
32. Figure 16. Given. \triangle ABC \cong \triangle DEF; M is the midpoint of \overline{BC} and N the midpoint of \overline{EF}. Prove. \overline{AM} \cong \overline{DN}.

Two stages: lengths and angles

33. In Figure 10, with the Given of Example 11, BE = 2x + 3 and DE = 5x - 9. Find x and BD.
34. In Figure 11, with the Given of Example 12, m\angle AEB = (3y + 15)^\circ. Find y.
35. In Figure 11, with the Given of Example 12, m\angle BAC = 35^\circ and m\angle BCA = 25^\circ. Find m\angle ADC and m\angle ABE.
36. In Figure 12, with the Given of Example 13, FE = 3t - 4 and GE = t + 6. Find t and FG.
37. In Figure 13, with the Given of Example 14, m\angle ADB = 130^\circ. Find m\angle BDE and m\angle CDE.
38. In Figure 13, with the Given of Example 14, BC = 18, DB = 15 and AB = 41. Find DE, AE and AD.
39. A kite ABCD has AB = AD = 17, CB = CD = 10 and diagonal BD = 16, and its diagonals \overline{AC} and \overline{BD} meet at E. Find AC.

Two stages: writing the proofs

40. Using Figure 10 and the Given of Example 11, prove \overline{AC} \perp \overline{BD}.
41. Using Figure 12, prove Exercise 30’s Prove from its Given.
42. Using Figure 13 without its tick marks and a new Given, \angle BAD \cong \angle CAD and \angle BDA \cong \angle CDA, prove \overline{BE} \cong \overline{CE}.
43. Using Figure 16 and the Given of Exercise 32, prove \overline{AM} \cong \overline{DN}. You may cite “a midpoint makes each half half as long as the whole”, proved in justifying each step.

Reasoning

44. Use Figure 7 without its arcs. Given. \overline{PM} is the perpendicular bisector of \overline{AB}. Prove. \overrightarrow{PM} bisects \angle APB. Then say what else the same congruence proves about P.
45. A student wants to prove \angle A \cong \angle D and plans to show \triangle ABC \cong \triangle DEF by ASA, using \angle A \cong \angle D as one of the three pairs. Explain what is wrong with the plan.
46. Two triangles have been proved congruent, and a proof needs a pair of segments that are not sides of either triangle. Can corresponding parts of these triangles supply the pair? Explain.
47. Explain why “corresponding parts of congruent triangles are congruent” needs no proof of its own, while the AAS theorem does.
48. In Example 9, suppose the observers swap the angles by mistake, laying out 50^\circ at A and 70^\circ at B on land. Is AT still the distance from A to the ship? Which distance does it equal?
49. After proving \triangle ABC \cong \triangle DEF by SAS from \overline{AB} \cong \overline{DE}, \angle B \cong \angle E and \overline{BC} \cong \overline{EF}, a proof cites corresponding parts for \overline{AB} \cong \overline{DE}. Is that wrong? Is it useful?
50. Quadrilaterals satisfy ABCD \cong PQRS. Does \angle B \cong \angle Q follow? Would four pairs of congruent sides have been enough to prove ABCD \cong PQRS? Explain the difference.
51. A problem states \triangle ABC \cong \triangle DEF as a given. Which criterion must a proof cite before it may claim \overline{BC} \cong \overline{EF}? Explain.
52. Given. \overline{AB} \cong \overline{AD} and \angle BAC \cong \angle DAC in the kite of Figure 10. Prove. \overline{CB} \cong \overline{CD}. Does this need two stages? Explain.
53. A proof of Example 11 begins: “1. \overline{AB} \cong \overline{AD} (given). 2. \overline{AE} \cong \overline{AE} (reflexive property). 3. \angle BAE \cong \angle DAE (the diagonal cuts the angle in half).” What is wrong with line 3, and how is it repaired?
54. In Example 14, the second stage could use \triangle ABE and \triangle ACE instead. Which part does stage one supply for them, under which name, and which criterion finishes?
55. Explain why the helper congruence must appear above the target congruence in a two-column proof, and never below it.