# R.D. Sharma Solutions Class 9th: Ch 10 Congruent Triangles Exercise 10.1

#### Chapter 10 Congruent Triangles R.D. Sharma Solutions for Class 9th Exercise 10.1

**Exercise 10.1**

1. In Fig. 10.22, the sides BA and CA have been produced such that BA = AD and CA = AE. Prove that segment DE || BC.

**Solution**

2. In a ΔPQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, OR, and RP respectively. Prove that LN = MN.

**Solution**

3. In Fig. 10.23, PQRS is a square and SRT is an equilateral triangle. Prove that

(i) PT = QT

(ii) ∠TQR = 15°

**Solution**

(ii)

**Solution**

5. In a Δ ABC, if ∠A = 120° and AB = AC. Find ∠B and ∠C.

**Solution**

**Solution**

7. The vertical angle of an isosceles triangle is 100°. Find its base angles.

**Solution**

8. In Fig. 10.24, AB = AC and ∠ACD = 105°, find ∠BAC.

**Solution**

9. Find the measure of each exterior angle of an equilateral triangle.

**Solution**

10. It is given that the base of an isosceles triangle is produced on both sides.

**Solution**

11. In Fig. 10.25, AB = AC and DB = DC, find the ratio ∠ABD : ∠ACD.

**Solution**

12. Determine the measure of each of the equal angles of a right-angled isosceles triangle.

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

**Solution**

It is given that,

ABC is right angled triangle where,

ABC is right angled triangle where,

∠A = 90°

AB = AC

We have to find ∠B and ∠C.

Since, AB = AC

∠B = ∠C (Isoceles triangle)

Now,

∠A +∠B+∠C = 180 (Property of triangle)

⇒ 90 + 2∠B = 180 (∠B = ∠C)

⇒ 2∠B = 180 - 90

⇒ ∠B = 45

So, ∠B = ∠C = 45

Hence,

∠B = 45° and ∠C = 45°

13. 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 (See Fig. 10.26). Show that the line PQ is perpendicular bisector of AB.

**Solution**