Congruent Triangles Problems: ASA, AAS and SSS

On this page we solve problems with three congruence criteria. Problems with the Side-Angle-Side criterion are on the page SAS problems, and the theory is on the page Congruence of Triangles.

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.

Angle-Angle-Side (AAS)

If two angles and a side opposite one of them in one triangle are equal to the corresponding two angles and side in another triangle, the triangles are congruent.

AAS follows from ASA: the third angles are equal too, because the angles of each triangle add up to ${180^\circ}.$

Side-Side-Side (SSS)

If the three sides of one triangle are equal to the three sides of another triangle, the triangles are congruent.

Problems

Problem 1

In the isosceles triangle $ABC$ we have ${AC=BC},$ and $AM$ and $BN$ are the bisectors of the base angles. Prove that ${AM=BN}.$

Solution. Consider the triangles $ABM$ and $BAN.$

1. $AB$ is a common side.
2. ${\angle ABM=\angle BAN},$ because the base angles of an isosceles triangle are equal.
3. ${\angle BAM=\angle ABN},$ because each of them is half of a base angle.

The side $AB$ lies between the two angles in both triangles, so by ASA ${\triangle ABM\cong\triangle BAN}.$ The segments $AM$ and $BN$ are corresponding sides, so ${AM=BN}.$

Problem 2

$CM$ is a median of triangle $ABC.$ The segments $AA_1$ and $BB_1$ are perpendicular to the line $CM$ (the points $A_1$ and $B_1$ lie on this line). Prove that ${AA_1=BB_1}.$

Solution. Consider the right triangles $AA_1M$ and $BB_1M.$

1. ${AM=BM},$ because $CM$ is a median.
2. ${\angle AMA_1=\angle BMB_1}$ as vertical angles.
3. ${\angle AA_1M=\angle BB_1M=90^\circ}.$

The equal sides $AM$ and $BM$ lie opposite the right angles, so by AAS ${\triangle AA_1M\cong\triangle BB_1M},$ and ${AA_1=BB_1}.$ (If ${CM\perp AB},$ the points $A_1,$ $B_1$ and $M$ coincide, and the statement is clear.)

Problem 3

Prove that every point of the bisector of an angle is equally far from the arms of the angle.

Solution. Let $M$ be a point of the bisector $OL$ of the angle $AOB,$ and let ${MP\perp OA}$ and ${MQ\perp OB}.$ Consider the triangles $OPM$ and $OQM.$

1. $OM$ is a common side.
2. ${\angle POM=\angle QOM},$ because $OL$ is the bisector.
3. ${\angle OPM=\angle OQM=90^\circ}.$

By AAS ${\triangle OPM\cong\triangle OQM},$ so ${MP=MQ}.$ The perpendiculars also cut equal segments from the arms: ${OP=OQ}.$

Problem 4

Prove that if the altitude and the angle bisector from the same vertex of a triangle coincide, the triangle is isosceles.

Solution. Let $CP$ be both the altitude and the bisector from $C.$ Consider the triangles $APC$ and $BPC.$

1. $CP$ is a common side.
2. ${\angle ACP=\angle BCP},$ because $CP$ is the bisector.
3. ${\angle APC=\angle BPC=90^\circ},$ because $CP$ is the altitude.

The side $CP$ lies between the two angles, so by ASA ${\triangle APC\cong\triangle BPC},$ and ${AC=BC}.$ The triangle is isosceles.

Problem 5

Prove that a triangle with two equal angles is isosceles: if ${\angle A=\angle B}$ in triangle $ABC,$ then ${AC=BC}.$

Solution. Draw the bisector $CD$ of the angle at $C.$ Consider the triangles $ACD$ and $BCD.$

1. ${\angle A=\angle B}$ (given).
2. ${\angle ACD=\angle BCD},$ because $CD$ is the bisector.
3. $CD$ is a common side.

The side $CD$ lies opposite the angle $A$ in the first triangle and opposite the equal angle $B$ in the second, so by AAS ${\triangle ACD\cong\triangle BCD}.$ Hence ${AC=BC}.$ Together with Problem 3 on the page SAS problems this shows that a triangle is isosceles exactly when it has two equal angles.

Problem 6

In the triangles $ABC$ and $A_1B_1C_1$ we know ${AB=A_1B_1},$ ${BC=B_1C_1},$ and the medians $AM$ and $A_1M_1$ to the sides $BC$ and $B_1C_1$ are equal. Prove that ${\triangle ABC\cong\triangle A_1B_1C_1}.$

Solution. $M$ and $M_1$ are midpoints, so ${BM}={\frac12\,BC}={\frac12\,B_1C_1}={B_1M_1}.$ In the triangles $ABM$ and $A_1B_1M_1\text{:}$

1. ${AB=A_1B_1};$
2. ${BM=B_1M_1};$
3. ${AM=A_1M_1}.$

By SSS ${\triangle ABM\cong\triangle A_1B_1M_1},$ so ${\angle B=\angle B_1}.$ Now the triangles $ABC$ and $A_1B_1C_1$ have ${AB=A_1B_1},$ ${\angle B=\angle B_1}$ and ${BC=B_1C_1},$ so by SAS ${\triangle ABC\cong\triangle A_1B_1C_1}.$

Problem 7

In the isosceles triangle $ABC$ we have ${AC=BC},$ and $M$ is the midpoint of the base $AB.$ Prove that ${CM\perp AB}$ and that $CM$ bisects the angle at $C.$

Solution. In the triangles $AMC$ and $BMC$ we have ${AC=BC},$ ${AM=BM}$ and the common side $CM.$ By SSS ${\triangle AMC\cong\triangle BMC}.$ So ${\angle ACM=\angle BCM},$ and ${\angle AMC=\angle BMC}.$ The last two angles are adjacent, with a sum of ${180^\circ},$ so each of them is ${90^\circ}.$

Problem 8

In the convex quadrilateral $ABCD$ the opposite sides are equal: ${AB=CD}$ and ${BC=DA}.$ Prove that ${\angle B=\angle D}$ and ${AB\parallel CD}.$

Solution. Draw the diagonal $AC.$ In the triangles $ABC$ and $CDA$ we have ${AB=CD},$ ${BC=DA}$ and the common side $AC.$ By SSS ${\triangle ABC\cong\triangle CDA}.$

So ${\angle B=\angle D},$ and also ${\angle BAC=\angle DCA}.$ These are alternate angles at the lines $AB$ and $CD$ cut by $AC,$ so ${AB\parallel CD}.$ In the same way ${BC\parallel AD}:$ the quadrilateral is a parallelogram.

Problem 9

In the quadrilateral $ABCD$ (a kite) ${AB=AD}$ and ${CB=CD}.$ Prove that the diagonal $AC$ bisects the angles at $A$ and at $C.$

Solution. In the triangles $ABC$ and $ADC$ we have ${AB=AD},$ ${CB=CD}$ and the common side $AC.$ By SSS ${\triangle ABC\cong\triangle ADC},$ so ${\angle BAC=\angle DAC}$ and ${\angle BCA=\angle DCA}.$

Problem 10

The segments $AB$ and $CD$ cross at the point $O,$ which is the midpoint of $AB,$ and ${AC\parallel BD}.$ Prove that $O$ is also the midpoint of $CD.$

Solution. Consider the triangles $AOC$ and $BOD.$

1. ${AO=BO},$ because $O$ is the midpoint of $AB.$
2. ${\angle OAC=\angle OBD}$ as alternate angles at the parallel lines $AC$ and $BD$ cut by $AB.$
3. ${\angle AOC=\angle BOD}$ as vertical angles.

The side $AO$ lies between the angles at $A$ and $O,$ and $BO$ between the angles at $B$ and $O.$ By ASA ${\triangle AOC\cong\triangle BOD},$ so ${OC=OD}.$

Practice Problems

Try them yourself, then open the solution.

1. Prove that in an isosceles triangle the altitudes to the legs are equal.

Let ${AC=BC},$ and let $AA_1$ and $BB_1$ be the altitudes to the legs. The right triangles $ABA_1$ and $BAB_1$ have the common hypotenuse $AB$ and the equal angles ${\angle ABA_1=\angle BAB_1}.$ By AAS they are congruent, so ${AA_1=BB_1}.$

2. Prove that congruent triangles have equal medians to corresponding sides.

If ${\triangle ABC\cong\triangle A_1B_1C_1}$ and $M,$ $M_1$ are the midpoints of $BC$ and $B_1C_1,$ then ${AB=A_1B_1},$ ${\angle B=\angle B_1}$ and ${BM=B_1M_1}.$ By SAS ${\triangle ABM\cong\triangle A_1B_1M_1},$ so ${AM=A_1M_1}.$

3. The point $M$ is equally far from the ends of the segment $AB,$ ${MA=MB}.$ Prove that $M$ lies on the perpendicular bisector of $AB.$

Let $O$ be the midpoint of $AB.$ The triangles $MOA$ and $MOB$ have ${MA=MB},$ ${OA=OB}$ and the common side $MO.$ By SSS they are congruent, so the equal adjacent angles $MOA$ and $MOB$ are right angles: $MO$ is the perpendicular bisector.

4. In the triangles $ABC$ and $DEF$ we know ${\angle A=\angle D},$ ${\angle C=\angle F}$ and ${AB=DE}.$ Are the triangles congruent?

Yes, by AAS: the sides $AB$ and $DE$ lie opposite the equal angles $C$ and $F.$

More: Congruence of Triangles · Problems: SAS · Triangles