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;
AB = AD
∠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.
∠CBO = ∠OAD
∠OCB = ∠ODA
Now consider ▵AOD and ▵BOC;
∠CBO = ∠OAD
AD = BC
∠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