# NCERT Solutions for Class 9 Maths Chapter 7 Triangles

Q. Prove that measure of each angle an equilateral triangle is 60°.

Ans. Given : DABC is an equilateral triangle.
To prove : ∠A = ∠B = ∠C = 60° Proof : In DABC,
AB = BC
∠C = ∠A ...(i)
(Opposite angles of equal sides)
AB = AC
∠C = ∠B ...(ii)
From equations (i) and (ii), ∠A = ∠B = ∠C ...(iii)
In △ABC,
∠A + ∠B + ∠C = 180° (Sum of all angles of △)
⇒ 3∠A = 180°
⇒ ∠A = 60°
∠A = ∠B = ∠C = 60°.

Q. In the given figure, l and m are two parallel lines intersected by another pair of parallel lines p and q. Show that △ABC △CDA. Ans. Given : l || m and p || q
To prove : △ABC △CDA
Proof : l || m
∠ACB = ∠DAC ...(i)
(Alternative angles)
p || q
∠BAC = ∠ACD ...(ii)
(Alternativ
e angles)
∠ACB = ∠DAC (Proved above)
∠BAC = ∠ACD (Proved above)
AC = AC (Common)
△ABC △CDA (ASA congruency rule)

Q. AB is a line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B. Show that the line PQ is perpendicular bisector of AB.

Ans. Give : P and Q are equidistant from the point A and B respectively. To prove : PQ is perpendicular bisector of AB
Proof : In DAPQ and DBPQ
AP = BP (Given)
AQ = BQ (Given)
PQ = PQ (Common)
APQ BPQ (SSS congruency rule)
∠APQ = ∠BPQ (CPCT)
AP = BP (Given)
OP = OP (Common)
AOP BOP (SAS congruency rule)
AO = BO (CPCT) ∠AOP = ∠BOP (CPCT)
∠AOP + ∠BOP = 180° (Linear pair)
∠AOP + ∠AOP = 180°
∠AOP = 90°
PQ ⊥ AB

Q. If the bisector of the vertical angle of a triangle bisects the base of a triangle then the triangle is isosceles.

Ans. Given : In DABC, AD is the angle bisector of ∠BAC and also bisects the base. To prove : ABC is an isosceles triangle.
CE
Proof : In ABD and DCE,
(Vertically opposite angles)
BD = DC (Given)
ABD ECD (SAS congruency rule)
AB = CE (CPCT) ...(i)
[From (ii) and (iii)]
In ACE, ∠CAE = ∠CEA
(Opposite sides of equal angles)
CE = AB [from (i)]
Form (iv) and (i),
AB = AC
ABC is an isosceles triangle.

Q. DABC is an isosceles triangle with AB = AC, side BA is produced to D such that AB = AD. Prove that ∠BCD is a right angle. Ans. Given : △ABC is an isosceles triangle and BA = AD. To prove : ∠BCD = 90°
Proof : AB = AC
∠ABC = ∠ACB ...(i)
(Opposite angles of equal sides)