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# $ \triangle \mathrm{ABC} $ is an isosceles triangle in which $ \mathrm{AB}=\mathrm{AC} $. Side $ \mathrm{BA} $ is produced to $ \mathrm{D} $ such that $ \mathrm{AD}=\mathrm{AB} $ (see Fig. 7.34). Show that $ \angle \mathrm{BCD} $ is a right angle.

"

**Given:**

$\triangle ABC$ is an isosceles triangle in which $AB=AC$. Side $BA$ is produced to $D$ such that $AD=AB$.

**To do:**

We have to show that $\angle BCD$ is a right angle.

**Solution:**

Let us consider $\triangle ABC$,

Given,

AB = AC

We know that,

The angles opposite to the equal sides are also equal.

This implies,

$\angle ACB = \angle ABC$

Now, let us consider $\triangle ACD$

Given,

AD = AB

We know that,

The angles opposite to the equal sides are also equal.

This implies,

$\angle ADC= \angle ACD$

We know that,

The sum of the interior angles of a triangle is always equal to $180^o$

This implies

In $\triangle ABC,$

$\angle CAB+\angle ACB+\angle ABC = 180^o$

Therefore,

$\angle CAB + 2\angle ACB=180^o$

This implies,

$\angle CAB = 180^o–2\angle ACB$.........(i)

Similarly,

In $\triangle ADC$,

$\angle CAD = 180^o – 2\angle ACD$......(ii)

also, as $BD$ is a straight line,

We get,

$\angle CAB + \angle CAD = 180^o$

By adding (i) and (ii) we get,

$\angle CAB+\angle CAD=180^o–2\angle ACB+180^o–2\angle ACD$

$180^o=360^o–2\angle ACB-2\angle ACD$

$2(\angle ACB+\angle ACD)=180^o$

$\angle BCD=90^o$.

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