# Congruent triangle questions – set 01

Given below are collection of questions related to congruent triangles.

All the questions are to the standard of grade 9.

Question 01
In below figure, AB = AD and EA = AC. Prove that line ED is parallel to BC.

Solution
Consider ▵ABC and ▵ABC;

∠BAC = ∠EAD ( vertically opposite angle)
EA = AC

By SAS congruency, ▵ABC and ▵ABC are congruent to each other.

Since both the triangle are congruent, we can write;
∠ABC = ∠EDA

Both the angles are also alternate interior angles.

Since both of them are equal it means that ED & BC are parallel lines intersected by transversal BD.

Hence Proved.

Question 02
In the below image, L, M & N are midpoint of sides PQ, QR & RP respectively. If side PQ = QR then prove that LN = MN.

Solution
PQ = QR ( given )

\mathtt{\frac{1}{2} PQ=\frac{1}{2} QR}

PL = MR (Since L & M are midpoints of PQ & QR)

Also it’s given that PQ = QR;

We know that angle opposite to equal sides are equal.

∠LPN = ∠NRM

Now consider ▵LPN and ▵MRN;

PL = MR
∠LPN = ∠NRM
PN = NR

By SAS congruency condition both triangles are congruent.

So we can write LN = MN

Hence Proved.

Question 03
Consider triangle ABC in which AB = AC and ∠BAC = 120 degree. Find the value of ∠B and ∠C.

Solution
We know that in triangle, angle opposite to equal sides are equal.

Since side AB = AC, their opposite angles will be equal.

∠ABC = ∠ACB = x degree

Now using angle sum property of triangle we can write;

∠ABC + ∠BCA + ∠BAC= 180

x + x + 120 = 180

2x = 180 – 120

2x = 60

x = 30 degree

Hence, both ∠B & ∠C measures 30 degree.

Question 04
In the below image line BC & AD are equal & parallel to each other.. Prove that line AB & CD bisect each other at point O.

Solution
Note that line BC & AD are parallel intersected by transversal AB & CD.

According to parallel line property, below angles are alternate interior angles.

∠OCB = ∠ODA

Now consider ▵AOD and ▵BOC;

∠OCB = ∠ODA

By ASA condition, both the triangles are congruent.

So we can say that;
AO = OB
DO = OC

Hence, lines AB & CD bisect at point O.

Question 05
Given below is isosceles triangle ABC in which AB = AC. Also line BD and CE are angle bisector of respective angles. Prove that BD = CE.

Solution
We know that in triangle, angle opposite to equal sides are equal.

Since AB = AC, we can write;

∠B = ∠C

\mathtt{\frac{1}{2} \angle B=\frac{1}{2} \angle C}\\\ \\ \mathtt{\angle DBC\ =\ \angle ECB}

Now consider ▵EBC and ▵DCB;

∠B = ∠C

BC = CB ( common side)

∠ECB = ∠DBC

By ASA congruency, both the triangles are congruent. So we can write;

BD = CE

Hence Proved.

Question 06
Given below are set of triangle in which ∠A = ∠D. Also side BC = FE are equal. Prove that both the triangle are congruent.

Solution
Considering triangle ABC and DEF

∠A = ∠D
∠B = ∠E = 90 degree
BC = EF

By AAS congruency condition, both the triangles are congruent.

Hence Proved.

Question 07
In below triangle AD is angle bisector of ∠EAC. Also AD is parallel to BC. Prove that triangle ABC is an isosceles triangle.

Solution
AD is angle bisector of ∠EAC, so we can write;

∠1 = ∠2

Since AD is parallel to BC and EB is transversal, we can write;

∠ 1 = ∠ 3 ( corresponding angle)

Since ∠1 = ∠ 2, the above expression can be written as;

∠ 2 = ∠ 3

Also since AD & BC are parallel lines, we can write;
∠ 2 = ∠ 4 (alternate interior angles)

Utilizing the above two expression;
∠3 = ∠4

We know that in triangle sides opposite to equal angles are equal.

Since ∠3 = ∠4; we can write;

AB = AC

Hence, the given triangle is an isosceles triangle.

Question 08
In the below triangle side PQ = PR and line ST is parallel to QR. Prove that PS = PT

Solution
We know that angle opposite to equal side of triangles are equal.

Since PQ = PR, we can write;

∠Q= ∠R (eq 1)

Line ST is parallel to QR intersected by transversal PQ & PR.

∠PST = ∠PQR (alternate interior angle)

∠PTS = ∠PRQ

Putting both the value in equation (1) we get;

∠PST = ∠PTS

In triangle side opposite to equal angle are equal.
Hence, PS = ST