### 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

Given that AD is bisector ofâˆ EAC and AD || BC.

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

âˆ ABD + âˆ BAD + âˆ ADB = 180Â°

âˆ ADB + âˆ ADC = 180Â°

So,

âˆ ABD + âˆ BAD + âˆ ADB = âˆ ADB + âˆ ADC

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°.