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# Class 9 RD Sharma Solutions – Chapter 10 Congruent Triangles- Exercise 10.3

• Last Updated : 29 Jul, 2021

### Question 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:

We have two right triangles ABC and DEF in which one side and acute angle of one are equal to the corresponding side and angles of the other, i.e.,

âˆ A = âˆ D

BC = EF

Prove: Î” ABC â‰… Î” DEF

Proof:

From Î” ABC and Î” DEF

âˆ  B = âˆ  E = 90Â° â€¦â€¦.(i)[Right triangle]

BC = EF â€¦â€¦.(ii)[Given]

âˆ A = âˆ D â€¦â€¦(iii)[Given]

By AAS congruence criterion,

Î” ABC â‰… Î” DEF

Hence, proved.

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

Solution:

In ABC be a triangle

From the above figure,

âˆ 1 = âˆ 2 [AD is a bisector of âˆ  EAC]

âˆ 1 = âˆ 3 [Corresponding angles]

and

âˆ 2 = âˆ 4 [alternative angle]

Also, we have âˆ 3 = âˆ 4

AB = AC

So the sides AB and AC are equal so

Î” ABC is an isosceles triangle.

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

Solution:

Let us considered Î” ABC is isosceles triangle

In which AB = AC and âˆ B = âˆ C

Given: âˆ A = 2(âˆ B + âˆ C)

Find: The value of âˆ A, âˆ B, and âˆ C

So, we have

âˆ A = 2(âˆ B + âˆ C)

âˆ A = 2(âˆ B + âˆ B) [âˆ B = âˆ C because Î” ABC is isosceles triangle]

âˆ A = 2(2 âˆ B)

âˆ A = 4(âˆ B)

As we know that sum of angles in an isosceles triangle = 180Â°

So, âˆ A + âˆ B + âˆ C = 180Â°

Now put the value of âˆ A and âˆ C, we get

4 âˆ B + âˆ B + âˆ B = 180Â°

6 âˆ B =180Â°

âˆ B = 30Â°

Here, âˆ B = âˆ C

âˆ B = âˆ C = 30Â°

And âˆ A = 4 âˆ B

âˆ A = 4 x 30Â° = 120Â°

Hence, the value of âˆ A = 120Â°, âˆ B = 30Â°, and âˆ C = 30Â°.

### Question 4. PQR is a triangle in which PQ = PR and 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:

In Î”PQR is a triangle

It is given that PQ = PR, and S, a line is drawn parallel to QR and intersecting PR at T

So, ST || QR.

Prove: PS = PT

As we know that PQ = PR, so the given â–³PQR is an isosceles triangle.

Hence, âˆ  PQR = âˆ  PRQ

âˆ  PST = âˆ  PQR and âˆ  PTS = âˆ  PRQ [Corresponding angles as ST parallel to QR]

âˆ  PQR = âˆ  PRQ

âˆ  PST = âˆ  PTS

So, In Î” PST,

âˆ  PST = âˆ  PTS

Therefore, Î” PST is an isosceles triangle.

So, PS = PT.

Hence, proved.

### Question 5. In an â–³ABC, 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 âˆ MOC = âˆ ABC.

Solution:

In â–³ABC,

It is given that AB = AC and the bisector of âˆ B and âˆ C intersect at O.

Prove: âˆ MOC = âˆ ABC

It is given that AB = AC

So the â–³ABC is an isosceles triangle

Hence

âˆ B = âˆ C

âˆ ABC = âˆ ACB

From the figure BO and CO are bisectors of âˆ ABC and âˆ ACB

So,

âˆ ABO = âˆ OBC = âˆ ACO = âˆ OCB = 1/2 âˆ ABC = 1/2 âˆ ACB ………..(i)

In â–³OBC

âˆ BOC + âˆ MOC = 180Â° ………(ii) [Straight line]

âˆ OBC + âˆ OCB + âˆ BOC = 180Â°[Sum of angles in an isosceles triangle = 180Â°]

From equation (ii)

âˆ OBC + âˆ OCB + âˆ BOC = âˆ BOC + âˆ MOC

âˆ OBC + âˆ OCB = âˆ MOC

2âˆ OBC = âˆ MOC

From Equation(i)

2(1/2 âˆ ABC) = âˆ MOC

âˆ ABC = âˆ MOC

Hence proved

### Question 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:

Given that P is a point on the bisector of âˆ ABC, and PQ || AB.

Prove: â–³BPQ is an isosceles triangle

It is given that BP is bisector of âˆ ABC

So, âˆ ABP = âˆ PBC ……….(i)

Also given that PQ || AB

So, âˆ BPQ = âˆ ABP ……….(ii) [alternative angles]

From equation (i) and (ii), we have

âˆ BPQ = âˆ PBC

or âˆ BPQ = âˆ PBQ

In â–³BPQ

âˆ BPQ = âˆ PBQ

Hence, proved â–³BPQ is an isosceles triangle.

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

Solution:

Prove: Each angle of an equilateral triangle is 60Â°.

Let us considered we have an equilateral triangle â–³ABC

So, AB = BC = CA

Take AB = BC

So, âˆ A = âˆ C …….(i) [Because opposite angles to equal sides are equal]

Take BC = AC

âˆ B = âˆ A ………(ii) [Because opposite angles to equal sides are equal]

From (i) and (ii), we get

âˆ A = âˆ B = âˆ C ………….(iii)

As we already know that the sum of angles in a triangle = 180Â°

So, âˆ A + âˆ B + âˆ C = 180Â°

From equation(iii), we get

âˆ A + âˆ A + âˆ A = 180Â°

3âˆ A = 180Â°

âˆ A = 60Â°

So, âˆ A = âˆ B = âˆ C = 60Â°

Hence Proved

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

Solution:

Given that in â–³ABC

âˆ A =âˆ B = âˆ C

Prove: â–³ABC is equilateral

In â–³ABC

As we already know that the sum of angles in a triangle = 180Â°

So, âˆ A + âˆ B + âˆ C = 180Â°

Given that âˆ A =âˆ B = âˆ C

So,

âˆ A + âˆ A + âˆ A = 180Â°

3âˆ A = 180Â°

âˆ A = 60Â°

So, âˆ A =âˆ B = âˆ C = 60Â°

As we know that the angles of equilateral triangles are of 60Â°

Hence, proved that â–³ABC is equilateral.

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

Solution:

Given that in â–³ABC,

âˆ B = 2 âˆ C, AD bisectors ofâˆ BAC, and AB = CD.

Prove:âˆ BAC = 72Â°

Now, draw a line BP which is bisector of âˆ ABC, and join PD.

Let us consideredâˆ C = âˆ ACB = y

âˆ B = âˆ ABC = 2y

Let us considered âˆ BAD = âˆ DAC = x

âˆ BAC = 2x [AD is the bisector of âˆ BAC]

In â–³BPC,

âˆ CBP = âˆ PCB = y [BP is the bisector of âˆ ABC]

PC = BP

In â–³ABP and â–³DCP,

âˆ ABP = âˆ DCP = y

AB = DC [Given]

And PC = BP

So, by SAS congruence criterion,

â–³ABP â‰… â–³DCP

âˆ BAP = âˆ CDP and AP = DP [C.P.C.T]

âˆ CDP = 2x

âˆ ADP = âˆ DAP = x

In â–³ABD

So,

2y + x = âˆ ADP + âˆ PDC

2y + x = x + 2x

2y = 2x

y = x

In â–³ABC,

As we already know that the sum of angles in a triangle = 180Â°

âˆ A + âˆ B + âˆ C = 180Â°

2x + 2y + y = 180Â° [âˆ A = 2x, âˆ B = 2y, âˆ C = y]

2(y) + 3y = 180Â° [x = y]

5y = 180Â°

y = 36Â°

âˆ A = âˆ BAC = 2 Ã— 36 = 72Â°

âˆ A = 72Â°

Hence proved

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

Solution:

Given that in ABC is a right-angled triangle

âˆ A = 90Â° and AB = AC

Find: âˆ B and âˆ C

In â–³ABC

AB = AC [Given]

âˆ B = âˆ C

As we already know that the sum of angles in a triangle = 180Â°

So, âˆ A + âˆ B + âˆ C = 180Â°

90Â° + âˆ B + âˆ B = 180Â°

2 âˆ B = 180Â°  – 90Â°

âˆ B = 45Â°

Hence, the value of âˆ B = âˆ C = 45Â°.

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