MENUCongruence of Triangles
Two triangles are congruent if one of them can be moved (slid, turned or flipped over) so that it exactly covers the other.
Congruent triangles have the same shape and the same size; only their position may differ. When one covers the other, each vertex lands on a vertex, each side on a side and each angle on an angle. These pairs are the corresponding vertices, sides and angles, and corresponding sides and angles are equal.
We write ${\triangle ABC\cong\triangle DEF}$ and read it "triangle $ABC$ is congruent to triangle $DEF$". The order of the letters matters: it shows which vertices correspond, $A$ to $D,$ $B$ to $E$ and $C$ to $F.$ So this short notation contains six equalities:
In proofs this is used as the rule "corresponding parts of congruent triangles are equal", often shortened to CPCTC.
On this page: Criteria · SAS · SSS · ASA · AAS · SSA · Right triangles · Writing a proof · Common mistakes · Practice
Congruence Criteria
To prove that two triangles are congruent, we do not need to check all six equalities: three well-chosen ones are enough. The rules that say which three are enough are the congruence criteria (also called congruence postulates and theorems). In the figures the given equal parts are red.
Side-Angle-Side (SAS)
If two sides and the angle between them in one triangle are equal to two sides and the angle between them in another triangle, the triangles are congruent.
If ${AB=A_1B_1},$ ${AC=A_1C_1}$ and ${\angle A=\angle A_1},$ then ${\triangle ABC\cong\triangle A_1B_1C_1}.$ The angle must be the one between the two sides, the included angle.
Side-Side-Side (SSS)
If the three sides of one triangle are equal to the three sides of another triangle, the triangles are congruent.
So three sides determine a triangle completely. That is why a frame of three sticks is rigid, while a frame of four sticks can be pushed out of shape. Bridges, roofs and cranes are built of triangles for this reason.
Angle-Side-Angle (ASA)
If a side and the two angles at its ends in one triangle are equal to a side and the two angles at its ends in another triangle, the triangles are congruent.
If ${AB=A_1B_1},$ ${\angle A=\angle A_1}$ and ${\angle B=\angle B_1},$ then ${\triangle ABC\cong\triangle A_1B_1C_1}.$
Angle-Angle-Side (AAS)
If two angles and a side that is not between them in one triangle are equal to the corresponding two angles and side in another triangle, the triangles are congruent.
This follows from ASA. When two angles of one triangle are equal to two angles of the other, the third angles are equal too, because each of them is ${180^\circ}$ minus the other two. In the figure ${\angle C=\angle C_1},$ so the side $BC$ with the angles $B$ and $C$ at its ends gives ASA. Make sure the sides correspond: the equal sides must lie opposite equal angles.
Two sides and an angle that is not between them (SSA)
Two sides and an angle that is not between them are, in general, not enough.
In the figure, the triangles $ABC$ and $ABC'$ have the common side $AB,$ the common angle at $A$ and ${BC=BC'}.$ Still, they are not congruent: the circle with center $B$ and radius $BC$ crosses the ray from $A$ at two points.
SSA does work when the given angle lies opposite the longer of the two given sides (or the two sides are equal). If ${AB=A_1B_1},$ ${BC=B_1C_1},$ ${\angle A=\angle A_1}$ and ${BC\ge AB},$ then ${\triangle ABC\cong\triangle A_1B_1C_1}.$
In the figure the angle at $A$ lies opposite the shorter side, ${BC\lt AB}.$ If $BC$ were at least as long as $AB,$ the circle would cross the ray from $A$ only once, and the triangle would be determined.
Congruence of right triangles
In right triangles the right angles are already equal, so two more pairs of equal parts are enough:
- the two legs (this is SAS);
- a leg and an acute angle (ASA or AAS);
- the hypotenuse and an acute angle (AAS);
- the hypotenuse and a leg, the HL criterion. This is the case of SSA that works, because the right angle lies opposite the hypotenuse, the longest side.
What is not enough
- Three angles (AAA). Triangles with equal angles have the same shape but can have different sizes. They are similar, not necessarily congruent.
- SSA with the angle opposite the shorter side, as in the figure above.
- Only two pairs of equal parts. For example, two equal sides allow any angle between them.
How to Write a Congruence Proof
- Name the two triangles, with the vertices in corresponding order.
- List three pairs of equal parts and give a reason for each: given, a common side, vertical angles, a property of the figure, and so on.
- Name the criterion (SAS, SSS, ASA, AAS or HL) and check that the parts are in the right places, for example that the angle in SAS is between the sides.
- Write the conclusion ${\triangle\ldots\cong\triangle\ldots}$ and then use the corresponding parts.
Example 1. The segments $AB$ and $CD$ bisect each other at the point $O.$ Prove that ${AC=BD}.$
Solution. Consider the triangles $AOC$ and $BOD.$
- ${AO=BO},$ because $O$ is the midpoint of $AB.$
- ${\angle AOC=\angle BOD}$ as vertical angles.
- ${CO=DO},$ because $O$ is the midpoint of $CD.$
The angle at $O$ lies between the two sides, so by SAS ${\triangle AOC\cong\triangle BOD}.$ The sides $AC$ and $BD$ correspond, so ${AC=BD}.$
Example 2. In triangles $ABC$ and $DEF$ we know ${AB=DE=5},$ ${BC=EF=7}$ and ${\angle B=\angle E=40^\circ}.$ Are the triangles congruent? Which angle of $DEF$ is equal to the angle $C?$
Solution. The angle $B$ lies between the sides $AB$ and $BC,$ and the angle $E$ between $DE$ and $EF.$ By SAS ${\triangle ABC\cong\triangle DEF},$ and the angle $C$ corresponds to the angle $F,$ so ${\angle C=\angle F}.$
Example 3. In triangles $ABC$ and $DEF$ we know ${\angle A=\angle D=50^\circ},$ ${\angle B=\angle E=60^\circ}$ and ${BC=DF}.$ Are the triangles congruent?
Solution. Not by these data. The side $BC$ lies opposite the ${50^\circ}$ angle, but $DF$ lies opposite the ${60^\circ}$ angle $E,$ so these sides do not correspond. The side that corresponds to $BC$ is $EF.$ The triangles have the same angles, so they are similar. But ${BC=DF\gt EF},$ because in triangle $DEF$ the side $DF$ lies opposite the larger angle. So triangle $ABC$ is larger than triangle $DEF,$ and they are not congruent.
Common Mistakes
| Mistake | Wrong | Right |
|---|---|---|
| The angle in SAS is not between the sides | ${AB=A_1B_1},$ ${BC=B_1C_1},$ ${\angle A=\angle A_1},$ so SAS | The angle $A$ is not between $AB$ and $BC;$ that is SSA, which may fail |
| Three angles used as a criterion | All angles are equal, so the triangles are congruent | They are similar; the sizes may differ |
| The order of the letters ignored | ${\triangle ABC\cong\triangle DEF},$ so ${AB=EF}$ | $AB$ corresponds to $DE,$ so ${AB=DE}$ |
| Sides that do not correspond | Two equal angles and one equal side anywhere give AAS | The equal sides must lie opposite equal angles |
Practice Problems
Try them yourself, then open the answer.
1. ${\triangle KLM\cong\triangle PQR}.$ Write the six equal pairs of sides and angles.
${KL=PQ},$ ${LM=QR},$ ${MK=RP},$ ${\angle K=\angle P},$ ${\angle L=\angle Q},$ ${\angle M=\angle R}.$
2. ${AB=DE},$ ${\angle A=\angle D},$ ${AC=DF}.$ Which criterion shows that ${\triangle ABC\cong\triangle DEF}?$
SAS: the angle $A$ is between the sides $AB$ and $AC.$
3. ${\angle A=\angle D},$ ${AB=DE},$ ${\angle B=\angle E}.$ Which criterion?
ASA: the side $AB$ lies between the angles $A$ and $B.$
4. ${\angle A=\angle D},$ ${\angle B=\angle E},$ ${BC=EF}.$ Which criterion?
AAS: $BC$ and $EF$ lie opposite the equal angles $A$ and $D.$
5. ${AB=DE},$ ${BC=EF},$ ${\angle A=\angle D}.$ Are the triangles always congruent?
Not always: this is SSA. They are congruent if the angle lies opposite the longer side (or the two sides are equal), that is if ${BC\ge AB}.$
6. Two right triangles have equal hypotenuses and one pair of equal legs. Are they congruent?
Yes, by the HL criterion.
7. The angles of two triangles are ${40^\circ},$ ${60^\circ},$ ${80^\circ}.$ Are the triangles congruent?
Not necessarily: they are similar, but one can be larger than the other.
8. $AB$ and $CD$ are diameters of a circle with center $O.$ Prove that ${AC=BD}.$
${OA=OB=OC=OD}$ as radii, and ${\angle AOC=\angle BOD}$ as vertical angles. By SAS ${\triangle AOC\cong\triangle BOD},$ so ${AC=BD}.$
Solved problems for each criterion: Side-Angle-Side and Angle-Side-Angle, Angle-Angle-Side and Side-Side-Side. More practice: problems on congruence of triangles.
More about triangles: Triangles · Classification · Altitude · Median · Similar triangles
