# Congruent triangle problems – set 04

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

∠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.

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;

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.