**Question 1:**

**In ΔABC, if AB = AC and ∠A = 70**

^{o}, find ∠B and ∠C.

**Solution:**

AB = AC implies their opposite angles are equal

But

**Question 2:**

**The vertical angle of an isosceles triangle is 100 ^{o}. Find its base angles.**

**Solution:**

Consider the isosceles triangle

Since the vertical angle of ABC is 100^{o}, we have, ^{o}

By angle sum property of a triangle, we have,

**Question 3:**

**In a ΔABC, if AB = AC and ∠B = 65**

^{o}, find ∠C and ∠A.

**Solution:**

In

Also by angle sum property, we have

**Question 4:**

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

**Solution:**

Let ABC be an isosceles triangle in which AB = AC.

Then we have

Let

Then vertex angle A = 2(x + x) = 4x

Now, x + x + 4x = 180^{o}

Therefore, Vertex

And,

**Question 5:**

**What is the measure of each of the equal angles of a right-angled isosceles triangle?**

**Solution:**

In a right angle isosceles triangle, the vertex angle is

Let x^{o} be the base angle and we have,

By angle sum property of a triangle, we have

Thus, we have

**Question 6:**

**If the base of an isosceles triangle is produced on both sides, prove that the exterior angles so formed are equal to each other.**

**Solution:**

Given: ABC is an isosceles triangle in which AB = AC and BC is produced both ways

To Prove:

Proof: In

AB = AC

Now exterior

and exterior

**Question 7:**

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

**Solution:**

Let

Since it is an equilateral triangle, all the angle is 60^{o}

The exterior angle of

The exterior angle of

The exterior angle of

We can observe that angles

Therefore, we have

Similarly, we have

Also, we have

Thus, we have,

So, the measure of each exterior angle of an equilateral triangle is 120^{o}.

**Question 8:**

**In the given figure, O is the mid-point of each of the line segments AB and CD. Prove that AC = BD and AC || BD.**

**Solution:**

Given: Two lines AB and CD intersect at O and O is the midpoint of AB and CD.

To prove: AC = BD and AC || BD

Proof:

In

AO = OB [Given: O is the midpoint of AB]

CO = OD [Given: O is the midpoint of CD]

So, by Side-Angle-Side congruence triangle are equal.

Therefore, we have, AC = BD.

Similarly, by c.p.c.t, we have,

This implies that alternate angles formed by AC and BD with transversal CD are equal. This means that, AC || BD. Thus, AC = BD and AC || BD.

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