In this chapter we will solve questions related to congruent triangles.

All the questions are to standard of grade 09 and are provided with detailed solution.

**Question 01**

In the below image, line AC bisects ∠A and ∠C. Prove that AB = AD and CB = CD.

**Solution**

Consider triangle ABC and ADC.

∠BAC = ∠DAC ( AC is angle bisector of ∠A )

AC = CA ( common side )

∠BCA = ∠DCA ( AC bisect ∠C)

By ASA congruency, both the triangles are congruent.

So, AB = AD and CB = CD.

Hence Proved.

**Question 02**

In the below image OA = OB and OQ = OP. Prove that PX = QX

**Solution**

OA = OB

OP + PA = OQ + QB

It’s given that OP = OQ, cancelling both of them from above expression.

PA = BQ

**Consider triangle OAQ and OBP;**

OA = OB ( given )

∠AOQ = ∠BOP ( Common angle )

OQ = OP ( given )

By SAS congruency, both the triangles are congruent.

So, ∠A = ∠B

**Now consider triangle XQB and XPA;**

∠BXQ = ∠AXP ( vertically opposite angle)

∠B = ∠A ( Proved above )

BQ = PA

By AAS congruency, both the triangles are congruent.

So, PX = QX

Hence Proved.

**Question 03**

In the below image quadrilateral ABCD is given in which AB || CD . P is the midpoint of side BC. Prove that AB = CQ.

**Solution**

Consider triangle ABP and QCP;

∠APB = ∠APC ( vertically opposite angles )

∠BAP = ** ** ∠CQP ( alternate interior angle)

BP = PC ( P is the midpoint of side BC)

By AAS congruency, both triangles are congruent.

So, AB = CQ

Hence Proved

**Question 04**

Given below is the triangle ABC in which AB = AC. In the figure if AX = AY then prove that CX = BY.

**Solution**

Consider triangle AXC and AYB;

AX = AY ( given )

∠XAC = ∠YAB ( common angle )

AC = AB ( given )

By SAS congruency, both triangles are congruent.

So, AX = AY

Hence Proved.

**Question 05**

Given below is triangle ABC in which D is midpoint of side BC. Line DL & DM are perpendicular on sides AB & AC respectively. If DL = DM, prove that side AB = AC.

**Solution****Consider triangle BLD & CMD;**

BD = DC ( given)

DL = DM ( given )

∠BLD = ∠CMD

By RHS congruency, both the triangles are congruent.

So, ∠B = ∠C

We now that in triangle, side opposite to equal angles are equal.

If, ∠B = ∠C then AB = AC.

Hence Proved.

**Question 06**

Given below is triangle ABC in which AB = AC. Also line OB & OC are angle bisector of ∠B and ∠C respectively. Prove that BO = CO and OA is bisector of angle A.

**Solution**

We know that angle opposite to equal sides are equal.

Since AB = AC, we can write ∠B = ∠C

Also BO & CO are angle bisectors, the above equation can also be expressed as;

\mathtt{\angle B\ =\ \angle C}\\\ \\ \mathtt{\frac{1}{2} \angle B=\frac{1}{2} \angle C}\\\ \\ \mathtt{\angle OBC\ =\ \angle OCB}

Consider triangle OBC, we know that side opposite to equal angle are equal.

Since ∠OBC = ∠OCB, we can write OB = OC

Consider triangle AOB & AOC;

AB = AC ( given )

OA = AO ( common side )

OB = OC (proved above)

By SSS congruency condition, both the triangles are congruent.

So, ∠ BAO = ∠CAO

Hence OA is bisector of ∠BAC.

**Question 07**

Given below is trapezium ABCD in which M & N are midpoint of side AB & CD. Also line MN is perpendicular to both side DM & CM. Prove that AD = BC

**Construction**

To solve the problems join DM and CM using dotted line.

**Solution****Consider triangle DNM and CNM**

DN = CN ( given )

NM = MN ( common side)

∠DNM = ∠CNM ( 90 degree)

By RHS congruency, both the triangles are congruent

So, **DM = CM and ∠DMN = ∠CMN**

We know that:

∠AMN = ∠BMN = 90 degree

Subtract ∠DMN on both sides;

∠AMN – ∠DMN = ∠BMN – ∠DMN

We know that; ∠DMN = ∠CMN

Putting this value in above equation.

∠AMN – ∠DMN = ∠BMN – ∠CMN

**∠AMD = ∠ BMC **

**Now consider triangle AMD & BMC;**

AM = MB ( M is midpoint)

∠AMD = ∠BMC ( proved above )

DM = CM ( proved above)

By SAS congruency, both triangles are congruent.

So, AD = BC

Hence Proved

**Question 08**

In below figure line BP is angle bisector of ∠B. If line QP is parallel to BA then prove that triangle BPQ is an isosceles triangle.

**Solution**

QP & BA are parallel lines intersected by BP as transversal.

∠1 = ∠QPB ( alternate interior angle )

Since BP is an angle bisector of ∠B, we can write;

∠1 = ∠2

Combining the two equation we get;

∠2 = ∠QPB

In a triangle, side opposite to equal angles are equal.

Since ∠2 = ∠QPB, so we can write QB = QP

Hence, QBP is an isosceles triangle.

**Question 09**

In the below figure, D is the midpoint of AC. Also BD = 1/2 AC. Show that ABC is a right triangle.

**Solution**

It’s given that AD = DC

Since BD = 1/2 AC, we can write following equation;

AD = DC = BD

Consider triangle ADB;

We know that angle opposite to equal sides are equal.

Since AD = DB, we can write ∠BAD = ∠DBA = x degree

Now consider triangle BDC;

Since BD = DC, we can write ∠DBC = ∠DCB = y degree

Now apply angle sum property in triangle ABC;

∠A + ∠B + ∠C = 180

x + ( x + y ) + y = 180

2x + 2y = 180

x + y = 90

∠B = x + y = 90 degree

Hence the given figure is of right triangle.