# $\triangle \mathrm{ABC}$ 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."

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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,

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,

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

$180^o=360^o–2\angle ACB-2\angle ACD$

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

$\angle BCD=90^o$.

Updated on 10-Oct-2022 13:41:11