# R.D. Sharma Solutions Class 9th: Ch 10 Congruent Triangles Exercise 10.3

#### Chapter 10 Congruent Triangles R.D. Sharma Solutions for Class 9th Exercise 10.3

**Exercise 10.3**

1. In two right triangles one side an acute angle of one are equal to the corresponding side and angle of the other. Prove that the triangles are congruent.

**Solution**

2. If the bisector of the exterior vertical angle of a triangle be parallel to the base. Show that the triangle is isosceles.

**Solution**

3. In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.

**Solution**

4. PQR is a triangle in which PQ = PR and S is any point on the side PQ. Through S, a line is drawn parallel to QR and intersecting PR at T. Prove that PS = PT.

**Solution**

5. In a ΔABC, it is given that AB = AC and the bisectors of ∠B and ∠C intersect at O. If M is a point on BO produced prove that ∠MOC = ∠ABC.

**Solution**

6. P is a point on the bisector of an angle ∠ABC. If the line through P parallel to AB meets BC at Q, prove that triangle BPQ is isosceles.

**Solution**

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

**Solution**

8. Angles A, B, C of a triangle ABC are equal to each other. Prove that ΔABC is equilateral.

**Solution**

9. ABC is a triangle in which ∠B = 2∠C. D is a point on BC such that AD bisects ∠BAC and AB = CD. Prove that ∠BAC = 72°.

**Solution**

∠B = 2∠C

AB = CD

AD bisects ∠BAC

To prove: ∠BAC = 72

Now, let ∠C = y then ∠B = 2y (Given)

Since, AD is a bisector of ∠BAC

So, let ∠BAD = ∠CAD = x

Let BP be the bisector of ∠ABC.

If we join PD we have,

In ΔBPC,

∠CBP = ∠BCP = y

So, BP = PC

In triangle ABP and DCP,

∠ABP = ∠DCP = y

AB = CD (Given)

BP = PC (Proved above)

So by SAS congruence criterion, we have ΔABP ≅ ΔADCP

⇒ ∠BAP = ∠CDP and AP = DP

∠CDP = 2x, and ∠ADP = ∠DAP = x (since ∠A = 2x)

In ΔABD,

we have ∠ADC = ∠ABD+ ∠BAC

Since, ∠ADC = ∠ADP+ ∠CDP

= x+2x = 3x

= x+2x = 3x

And, ∠ADC = ∠BAD+ ∠ABD = x+2y

So,

3x = x+2y

So,

3x = x+2y

⇒ 2x = 2y

⇒ x= y

In ABC,

we have ∠A+ ∠B+ ∠C = 180°

⇒ 2x + 2y + y = 180°

⇒ 5x = 180°

⇒ x = 36°

⇒ 5x = 180°

⇒ x = 36°

Here,∠BAC = 2x = 2×36 = 72°

10. ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.

**Solution**

It is given that,

∠A = 90°

AB = AC

We have to find ∠B and ∠C.

Since, AB = AC, therefore ∠B=∠C

Now, ∠A+ ∠B+ ∠C = 180° (property of triangle)

⇒ 90° + ∠B+ ∠B = 180° (Since ∠B= ∠C)

⇒ 90° +2∠B = 180°

⇒ 2∠B = 90°

⇒ ∠B = 90°/2

⇒ ∠B = 45°

Here, ⇒ ∠B = ∠C = 45°

Then, ∠A = 90°

Hence,

∠B = 45°

∠C = 45°