A congruence proof is a chain of justified statements. The drawing helps you find the triangles, but the proof must say why their matching parts agree. “They look the same” cannot establish a length, a right angle or a symmetry that the problem never gave.
Start by listing the givens separately from the required conclusion. Then match vertices, choose a valid test and use the congruence to justify a remaining corresponding part. All examples below describe non-degenerate triangles, and every equality needed for the initial test is stated explicitly.
Choose a test that fits the evidence
| Test | Required matching information | Important check |
|---|---|---|
| SSS | Three pairs of corresponding equal sides | All three pairs must match consistently |
| SAS | Two pairs of equal sides and their included angles | The angle must lie between the named sides |
| ASA or AAS | Two pairs of equal angles and one corresponding equal side | Match the side to the same vertices |
| RHS | Two right triangles with equal hypotenuses and one pair of equal corresponding legs | Identify the hypotenuse opposite each right angle |
Two equal angle pairs force the third pair to be equal because triangle angles total 180°. The additional matching side in ASA or AAS fixes the size. Equal angles alone do not do so. Likewise, two sides and a non-included angle are not a general congruence test.
Case 1: shared-side SSS without assuming the conclusion
Given AB = AC = 5 cm, BD = CD = 3 cm, and B, D, C on one straight line with D between B and C, prove that AD bisects angle BAC. Then show that AD is perpendicular to BC. No right angle at D is given.
| Statement | Reason |
|---|---|
| AB = AC | Given |
| BD = CD | Given |
| AD = AD | Same shared side |
| Triangle ABD ≅ triangle ACD | SSS, with A ↔ A, B ↔ C and D ↔ D |
| Angle BAD = angle CAD | Corresponding angles of congruent triangles |
| AD bisects angle BAC | Definition of an angle bisector |
In prose: triangles ABD and ACD have the equal side pairs AB and AC, BD and CD, and the common side AD. They are congruent by SSS. Their angles at A therefore agree, so AD divides angle BAC into two equal angles.
For the second conclusion, angle ADB equals angle ADC by the same correspondence. These adjacent angles total 180° because B, D and C are collinear. Two equal angles totalling 180° are each 90°, so AD is perpendicular to BC. Using that perpendicularity as an initial premise would be circular reasoning.
The givens are realizable: coordinates A(0, 4), B(−3, 0), C(3, 0), D(0, 0) produce sides 5, 5, 3 and 3. That independent coordinate check confirms the example is possible; the proof itself uses the stated side equalities.
Case 2: SAS and the included angle
For triangles PQR and XYZ, suppose PQ = XY = 6 cm, PR = XZ = 5 cm and angle QPR = angle YXZ = 60°. Prove QR = YZ and angle PQR = angle XYZ.
| Statement | Reason |
|---|---|
| PQ = XY and PR = XZ | Given side pairs |
| Angle QPR = angle YXZ | Given included angles |
| Triangle PQR ≅ triangle XYZ | SAS |
| QR = YZ | Corresponding sides of congruent triangles |
| Angle PQR = angle XYZ | Corresponding angles of congruent triangles |
In prose: the two known sides meet at P in the first triangle and X in the second. Their included angles agree, so SAS establishes congruence in the order PQR to XYZ. The remaining side QR corresponds to YZ, and the angle at Q corresponds to the angle at Y.
Do not write triangle PQR ≅ triangle XZY. That would match the 6 cm side PQ to the 5 cm side XZ. Correct facts can be spoiled by an incorrect vertex order. As a separate numerical check, the unlabelled third side in each diagram has squared length 6² + 5² − 2 × 6 × 5 × cos 60° = 31, consistent with the same side length.
Case 3: right angle, hypotenuse and side
Triangles MNO and UVW have right angles at M and U. Given MN = UV = 8 cm and NO = VW = 10 cm, prove MO = UW. The side opposite the right angle is the hypotenuse, so those hypotenuses are NO and VW.
| Statement | Reason |
|---|---|
| Angle NMO = angle VUW = 90° | Given right angles |
| NO = VW | Given equal hypotenuses |
| MN = UV | Given equal corresponding legs |
| Triangle MNO ≅ triangle UVW | RHS, with M ↔ U, N ↔ V, O ↔ W |
| MO = UW | Corresponding sides of congruent triangles |
In prose: both triangles are right-angled, their hypotenuses are equal and one pair of corresponding legs is equal. RHS therefore gives congruence, and the remaining legs match. A numerical check gives MO² = UW² = 10² − 8² = 36, so both lengths are 6 cm. The square root is positive because these are lengths.
Explain why incomplete evidence is insufficient
A learner argues that two equilateral triangles are congruent because each has angles 60°, 60°, 60°. Take one with side 4 cm and another with side 7 cm. Both satisfy the angle information, but their sides differ. This explicit counterexample disproves the claimed AA or AAA congruence rule.
Another learner sees two side equalities and any equal angle, then writes SAS. Ask which two sides form that angle. If it is opposite one of them, it is non-included, and the standard SAS test does not apply. A drawing may accidentally show congruent triangles, but the stated evidence alone has not established that conclusion. A matching included angle or a matching third side would provide a valid route.
Three solved practice tasks
- AB = DE = 3 cm, BC = EF = 4 cm, AC = DF = 5 cm. Prove angle ABC = angle DEF. The side pairs match A ↔ D, B ↔ E, C ↔ F. Thus triangle ABC ≅ triangle DEF by SSS, so their angles at B and E are equal.
- Angle A = angle D = 50°, angle B = angle F = 70°, and AB = DF = 9 cm. Match triangles ABC and DEF and find angle E. The included side gives ASA with triangle ABC ≅ triangle DFE. Thus C corresponds to E. Angle C = 180° − 50° − 70° = 60°, so angle E = 60°.
- Triangles JKL and MNP have angle J = angle M = 40° and angle K = angle N = 65°. Does congruence follow? No. Both third angles are 75°, establishing equal-angle shape but not size. A further equality such as JK = MN would fix a corresponding side and allow ASA; without it, congruence is unproved.
Read the final conclusion back through the ordered triangle names. Every equality should either be a given, an established fact or a consequence of the named test. This small discipline makes a proof clear enough to check without trusting the picture.
Frequently asked questions
What does congruent mean?
Congruent figures have the same shape and size. They may be turned, moved or reflected. Corresponding sides and corresponding angles are equal.
Why does the order of triangle names matter?
It records the vertex correspondence. In triangle ABC ≅ triangle PQR, A matches P, B matches Q and C matches R. Changing that order can make a later side or angle conclusion false.
Can a shared side count as evidence?
Yes. A segment has the same length as itself, so a shared side supplies one equality. Write its endpoints carefully, especially when the triangles name that side in opposite directions.
What is an included angle?
It is the angle between the two sides being used. If the sides are PQ and PR, the included angle has vertex P and is written angle QPR or RPQ.
Why is AA insufficient for congruence?
Two equal angles determine a triangle’s shape but leave its size free. Equilateral triangles of sides 4 cm and 7 cm have identical angles but are not congruent.
Why is RHS allowed when general SSA is not?
RHS also supplies a right angle and identifies the hypotenuse. A hypotenuse and one leg fix the other positive leg through Pythagoras, removing the ambiguity that can occur with a non-right SSA configuration.
May I measure the drawing to finish a proof?
Measured appearance is not deductive evidence. Use stated givens, marked information, definitions and previously established results. A diagram can guide your choice of triangles without supplying an unstated equality.
Further reading
For the underlying mathematics, see AMSI’s congruence tests and corresponding-part reasoning. The mathematics topic-test collection is a separate resource page that currently shows a member login; specific files and their answers are not verified here.
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