# NCERT Solutions for Class 9 Maths Chapter 7: Triangles

## NCERT Solutions for Class 9 Mathematics Chapter 7 Free PDF Download

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**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)

In △ABC and △ADC,

∠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.**Construction :** Extend AD such that AD = DE and join

CE**Proof :** In **△**ABD and **△**DCE,

∠ADB = ∠CDE

(Vertically opposite angles)

BD = DC (Given)

AD = DE (By construction)**△**ABD **≅****△**ECD (SAS congruency rule)

AB = CE (CPCT) ...(i)

∠BAD = ∠CED (CPCT) ...(ii)

∠BAD = ∠CAD (Given) ...(iii)

∠BAD = ∠CAD = ∠CED

[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)

AD = AC

∠ACD = ∠ADC ...(ii)

(Opposite angles of equal sides)

∠CAD = ∠ABC + ∠ACB

(Exterior angle property)

⇒ ∠CAD = 2∠ACB (from (i)) ...(iii)

∠BAC = ∠ADC + ∠ACD

(Exterior angle property)

⇒ ∠BAC = 2∠ACD (from (ii)) ...(iv)

From (iii) and (iv)

2∠ACB + 2∠ACD = ∠CAD + ∠BAC

⇒ 2(∠ACB + ∠ACD) = ∠BAD

⇒ 2(∠BCD) = 180°

⇒ ∠BCD = 90°.