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

**Exercise 10.3**

1.Â 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 Î”ABC, it is given that AB = AC and the bisectors of âˆ B and âˆ C intersect at O. If M is a point on BO produced prove that âˆ MOC = âˆ ABC.

**Solution**

6. P is a point on the bisector of an angle âˆ 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 60Â°

**Solution**

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

**Solution**

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

**Solution**

âˆ B = 2âˆ C

AB = CD

AD bisects âˆ BAC

To prove: âˆ BAC = 72

Now, let âˆ C = y then âˆ B = 2y (Given)

Since, AD is a bisector of âˆ BAC

So, let âˆ BAD = âˆ CAD = x

Let BP be the bisector of âˆ ABC.

If we join PD we have,

In Î”BPC,

âˆ CBP = âˆ BCP = y

So, BP = PC

In triangle ABP and DCP,

âˆ ABP = âˆ DCP = y

AB = CD (Given)

BP = PC (Proved above)

So by SAS congruence criterion, we have Î”ABP â‰…Â Î”ADCP

â‡’ âˆ BAP = âˆ CDP and AP = DP

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

In Î”ABD,

we have âˆ ADC = âˆ ABD+ âˆ BAC

Since, âˆ ADC = âˆ ADP+ âˆ CDP

= x+2x = 3x

= x+2x = 3x

And, âˆ ADC = âˆ BAD+ âˆ ABD = x+2y

So,

3x = x+2y

So,

3x = x+2y

â‡’ 2x = 2y

â‡’ x= y

In ABC,

we have âˆ A+ âˆ B+ âˆ C = 180Â°

â‡’ 2x + 2y + y = 180Â°

â‡’ 5x = 180Â°

â‡’ x = 36Â°

â‡’ 5x = 180Â°

â‡’ x = 36Â°

Here,âˆ BAC = 2x = 2Ã—36 = 72Â°

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

**Solution**

It is given that,

âˆ A = 90Â°

AB = AC

We have to find âˆ B and âˆ C.

Since, AB = AC, therefore âˆ B=âˆ C

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

â‡’ 90Â° + âˆ B+ âˆ B = 180Â° (Since âˆ B= âˆ C)

â‡’ 90Â° +2âˆ B = 180Â°

â‡’ 2âˆ B = 90Â°

â‡’ âˆ B = 90Â°/2

â‡’ âˆ B = 45Â°

Here, â‡’ âˆ B = âˆ C = 45Â°

Then, âˆ A = 90Â°

Hence,

âˆ B = 45Â°

âˆ C = 45Â°