### 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.,

∠A = ∠D

BC = EF

Prove: Δ ABC ≅ Δ DEF

Proof:

From Δ ABC and Δ DEF

∠ B = ∠ E = 90° …….(i)[Right triangle]

BC = EF …….(ii)[Given]

∠A = ∠D ……(iii)[Given]

By AAS congruence criterion,

Δ ABC ≅ Δ 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 of∠EAC and AD || BC.

From the above figure,

∠1 = ∠2 [AD is a bisector of ∠ EAC]

∠1 = ∠3 [Corresponding angles]

and

∠2 = ∠4 [alternative angle]

Also, we have ∠3 = ∠4

AB = AC

So the sides AB and AC are equal so

Δ 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 Δ ABC is isosceles triangle

In which AB = AC and ∠B = ∠C

Given: ∠A = 2(∠B + ∠C)

Find: The value of ∠A, ∠B, and ∠C

So, we have

∠A = 2(∠B + ∠C)

∠A = 2(∠B + ∠B) [∠B = ∠C because Δ ABC is isosceles triangle]

∠A = 2(2 ∠B)

∠A = 4(∠B)

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

So, ∠A + ∠B + ∠C = 180°

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

4 ∠B + ∠B + ∠B = 180°

6 ∠B =180°

∠B = 30°

Here, ∠B = ∠C

∠B = ∠C = 30°

And ∠A = 4 ∠B

∠A = 4 x 30° = 120°

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

### 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 ΔPQR 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 △PQR is an isosceles triangle.

Hence, ∠ PQR = ∠ PRQ

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

∠ PQR = ∠ PRQ

∠ PST = ∠ PTS

So, In Δ PST,

∠ PST = ∠ PTS

Therefore, Δ PST is an isosceles triangle.

So, PS = PT.

Hence, proved.

### Question 5. In an △ABC, 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:**

In △ABC,

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

Prove: ∠MOC = ∠ABC

It is given that AB = AC

So the △ABC is an isosceles triagnle

Hence

∠B = ∠C

∠ABC = ∠ACB

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

So,

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

In △OBC

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

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

From equation (ii)

∠OBC + ∠OCB + ∠BOC = ∠BOC + ∠MOC

∠OBC + ∠OCB = ∠MOC

2∠OBC = ∠MOC

From Equation(i)

2(1/2 ∠ABC) = ∠MOC

∠ABC = ∠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 ∠ABC, and PQ || AB.

Prove: △BPQ is an isosceles triangle

It is given that BP is bisector of ∠ABC

So, ∠ABP = ∠PBC ……….(i)

Also given that PQ || AB

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

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

∠BPQ = ∠PBC

or ∠BPQ = ∠PBQ

In △BPQ

∠BPQ = ∠PBQ

Hence, proved △BPQ is an isosceles triangle.

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

**Solution:**

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

Let us considered we have an equilateral triangle △ABC

So, AB = BC = CA

Take AB = BC

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

Take BC = AC

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

From (i) and (ii), we get

∠A = ∠B = ∠C ………….(iii)

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

So, ∠A + ∠B + ∠C = 180°

From equation(iii), we get

∠A + ∠A + ∠A = 180°

3∠A = 180°

∠A = 60°

So, ∠A = ∠B = ∠C = 60°

Hence Proved

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

**Solution: **

Given that in △ABC

∠A =∠B = ∠C

Prove: △ABC is equilateral

In △ABC

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

So, ∠A + ∠B + ∠C = 180°

Given that ∠A =∠B = ∠C

So,

∠A + ∠A + ∠A = 180°

3∠A = 180°

∠A = 60°

So, ∠A =∠B = ∠C = 60°

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

Hence, proved that △ABC is equilateral.

### Question 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:**

Given that in △ABC,

∠B = 2 ∠C, AD bisectors of∠BAC, and AB = CD.

Prove:∠BAC = 72°

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

Let us considered∠C = ∠ACB = y

∠B = ∠ABC = 2y

Let us considered ∠BAD = ∠DAC = x

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

In △BPC,

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

PC = BP

In △ABP and △DCP,

∠ABP = ∠DCP = y

AB = DC [Given]

And PC = BP

So, by SAS congruence criterion,

△ABP ≅ △DCP

∠BAP = ∠CDP and AP = DP [C.P.C.T]

∠CDP = 2x

∠ADP = ∠DAP = x

In △ABD

∠ABD + ∠BAD + ∠ADB = 180°

∠ADB + ∠ADC = 180°

So,

∠ABD + ∠BAD + ∠ADB = ∠ADB + ∠ADC

2y + x = ∠ADP + ∠PDC

2y + x = x + 2x

2y = 2x

y = x

In △ABC,

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

∠A + ∠B + ∠C = 180°

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

2(y) + 3y = 180° [x = y]

5y = 180°

y = 36°

∠A = ∠BAC = 2 × 36 = 72°

∠A = 72°

Hence proved

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

**Solution:**

Given that in ABC is a right-angled triangle

∠A = 90° and AB = AC

Find: ∠B and ∠C

In △ABC

AB = AC [Given]

∠B = ∠C

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

So, ∠A + ∠B + ∠C = 180°

90° + ∠B + ∠B = 180°

2 ∠B = 180° – 90°

∠B = 45°

Hence, the value of ∠B = ∠C = 45°.