Congruent Triangles Problems: Side-Angle-Side (SAS)

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.

In every problem below we name the two triangles, list three pairs of equal parts with a reason for each, and check that the angle lies between the two sides. The theory is on the page Congruence of Triangles, and problems for the other criteria are on the page ASA, AAS and SSS.

Problem 1

In the isosceles triangle $ABC$ we have ${AC=BC},$ and $CD$ is the bisector of the angle at $C$ (the point $D$ lies on $AB$). Prove that ${\triangle ACD\cong\triangle BCD}.$

Solution. In the triangles $ACD$ and $BCD\text{:}$

1. ${AC=BC},$ because the triangle is isosceles;
2. ${\angle ACD=\angle BCD},$ because $CD$ is the angle bisector;
3. $CD$ is a common side.

The angle $ACD$ lies between the sides $AC$ and $CD,$ and the angle $BCD$ lies between $BC$ and $CD.$ By SAS ${\triangle ACD\cong\triangle BCD}.$

Problem 2

Prove that in an isosceles triangle the bisector of the vertex angle is also the median and the altitude to the base.

Solution. Use the figure and the result of Problem 1: ${\triangle ACD\cong\triangle BCD}.$

• The sides $AD$ and $BD$ correspond, so ${AD=BD}:$ the point $D$ is the midpoint of $AB,$ and $CD$ is the median.
• The angles $ADC$ and $BDC$ correspond, so they are equal. They are also adjacent angles with a sum of ${180^\circ},$ so each of them is ${90^\circ}:$ $CD$ is the altitude.

From a vertex there is only one angle bisector, one median and one altitude, so in an isosceles triangle these three segments to the base coincide.

Problem 3

Prove that the base angles of an isosceles triangle are equal.

Solution. Draw the bisector $CD$ of the vertex angle. By Problem 1, ${\triangle ACD\cong\triangle BCD}.$ The angles at $A$ and $B$ are corresponding angles of these triangles, so ${\angle A=\angle B}.$

Problem 4

In the isosceles triangle $ABC$ we have ${AC=BC},$ and $D$ is the midpoint of the base $AB.$ The perimeter of triangle $ADC$ is 18 cm and ${CD=6}$ cm. Find the perimeter of triangle $ABC.$

Solution. The perimeter of $ADC$ is ${AC+CD+AD=18},$ so ${AC+AD=18-6=12}$ cm. Since ${BC=AC}$ and ${DB=AD},$ the perimeter of $ABC$ is

${AC+BC+AD+DB}={2\,(AC+AD)}={2\cdot 12}={24}$ cm.

Problem 5

A line cuts equal segments ${OC=OD}$ from the arms of the angle $AOB.$ Prove that the line is perpendicular to the bisector of the angle.

Solution. Let the line ${\ell=CD}$ cross the bisector $OL$ at the point $F.$ The triangle $OCD$ is isosceles, because ${OC=OD},$ and $OF$ is the bisector of its vertex angle. By Problem 2, $OF$ is also the altitude to the base $CD,$ so ${CD\perp OL}.$

Problem 6

The diagonals of the quadrilateral $ABCD$ bisect each other at the point $O.$ Prove that its opposite sides are equal.

Solution. In the triangles $AOD$ and $COB\text{:}$

1. ${AO=CO}$ and ${OD=OB},$ because the diagonals bisect each other;
2. ${\angle AOD=\angle COB}$ as vertical angles.

By SAS ${\triangle AOD\cong\triangle COB},$ so ${AD=CB}.$ In the same way ${\triangle AOB\cong\triangle COD},$ so ${AB=CD}.$

The congruence also gives ${\angle OAD=\angle OCB}.$ These are alternate angles at the lines $AD$ and $BC$ cut by $AC,$ so ${AD\parallel BC};$ in the same way ${AB\parallel CD}.$ So a quadrilateral whose diagonals bisect each other is a parallelogram.

Problem 7

${\triangle ABC\cong\triangle A_1B_1C_1}.$ The points $M$ and $M_1$ lie on the sides $AB$ and $A_1B_1,$ and ${AM=A_1M_1}.$ Prove that ${CM=C_1M_1}$ and ${\angle BMC=\angle B_1M_1C_1}.$

Solution. From the congruence of the triangles, ${AC=A_1C_1}$ and ${\angle A=\angle A_1}.$ In the triangles $AMC$ and $A_1M_1C_1\text{:}$

1. ${AM=A_1M_1}$ (given);
2. ${\angle A=\angle A_1};$
3. ${AC=A_1C_1}.$

By SAS ${\triangle AMC\cong\triangle A_1M_1C_1},$ so ${CM=C_1M_1}$ and ${\angle AMC=\angle A_1M_1C_1}.$ The angles $BMC$ and $B_1M_1C_1$ are adjacent to these equal angles, so ${\angle BMC}={180^\circ-\angle AMC}={180^\circ-\angle A_1M_1C_1}={\angle B_1M_1C_1}.$

Problem 8

The points $D$ and $E$ lie on the base $AB$ of the isosceles triangle $ABC$ (${AC=BC}$), and ${AD=BE}.$ Prove that ${CD=CE}.$

Solution. In the triangles $ADC$ and $BEC\text{:}$

1. ${AD=BE}$ (given);
2. ${\angle A=\angle B},$ the base angles (Problem 3);
3. ${AC=BC}.$

By SAS ${\triangle ADC\cong\triangle BEC},$ so ${CD=CE}.$

Problem 9

In the isosceles triangle $ABC$ (${AC=BC}$), $AA_1$ and $BB_1$ are the medians to the legs. Prove that ${AA_1=BB_1}.$

Solution. In the triangles $ABB_1$ and $BAA_1\text{:}$

1. $AB$ is a common side;
2. ${\angle BAB_1=\angle ABA_1},$ the base angles;
3. ${AB_1}={\frac12\,AC}={\frac12\,BC}={BA_1}.$

By SAS ${\triangle ABB_1\cong\triangle BAA_1},$ so ${BB_1=AA_1}.$

Problem 10

The squares $BMNC$ and $CPQA$ are built outwards on the sides $BC$ and $CA$ of triangle $ABC,$ whose angle $\gamma$ at $C$ is acute. Prove that ${AN=BP}.$

Solution. Consider the triangles $NCA$ and $BCP$ (dashed in the figure).

1. ${CN=CB},$ as sides of the square $BMNC;$
2. ${CA=CP},$ as sides of the square $CPQA;$
3. ${\angle NCA=90^\circ+\gamma}$ and ${\angle BCP=\gamma+90^\circ},$ so ${\angle NCA=\angle BCP}.$

By SAS ${\triangle NCA\cong\triangle BCP},$ so ${AN=BP}.$

In fact, the turn by ${90^\circ}$ around $C$ that takes $N$ to $B$ also takes $A$ to $P.$ So it takes the segment $NA$ to $BP,$ and therefore ${AN\perp BP}$ as well. This argument does not use the size of $\gamma,$ so ${AN=BP}$ for an obtuse angle at $C$ too.

Practice Problems

Try them yourself, then open the solution.

1. The points $A$ and $C$ lie on one arm of an angle with vertex $O,$ and $B$ and $D$ on the other, with ${OA=OB}$ and ${OC=OD}.$ Prove that ${AD=BC}.$

In the triangles $OAD$ and $OBC\text{:}$ ${OA=OB},$ the angle $O$ is common, and ${OD=OC}.$ By SAS they are congruent, so ${AD=BC}.$

2. $ABCD$ is a square, $M$ is the midpoint of $AB$ and $N$ is the midpoint of $AD.$ Prove that ${CM=CN}.$

In the triangles $CBM$ and $CDN\text{:}$ ${CB=CD},$ ${\angle B=\angle D=90^\circ}$ and ${BM=DN}$ (halves of equal sides). By SAS they are congruent, so ${CM=CN}.$

3. In triangle $ABC$ the sides ${AB=AC}.$ The points $D$ on $AB$ and $E$ on $AC$ are such that ${AD=AE}.$ Prove that ${BE=CD}.$

In the triangles $ABE$ and $ACD\text{:}$ ${AB=AC},$ the angle $A$ is common, and ${AE=AD}.$ By SAS they are congruent, so ${BE=CD}.$

4. The points $D,$ $E$ and $F$ lie on the sides $AB,$ $BC$ and $CA$ of the equilateral triangle $ABC,$ with ${AD=BE=CF}.$ Prove that the triangle $DEF$ is equilateral.

${BD=AB-AD},$ ${CE=BC-BE}$ and ${AF=CA-CF}$ are equal too. The triangles $ADF,$ $BED$ and $CFE$ each have a ${60^\circ}$ angle between two sides of these equal lengths, so by SAS they are congruent. Hence ${DF=ED=FE}.$

More: Congruence of Triangles · Problems: ASA, AAS and SSS · Triangles