In the adjoining figure, $ \Delta A B C $ is an isosceles triangle such that $ A B=A C $. Side BA is produced to D such that $AD=AB$. Show that $ \angle B C D=90^{\circ} $.
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Given:

 \( \Delta A B C \) is an isosceles triangle such that \( A B=A C \). Side BA is produced to D such that $AD=AB$.

To do:

We have to show that \( \angle B C D=90^{\circ} \).

Solution:

In $\vartriangle ABC$,

$AB=AC$

This implies,

$\angle ACB=\angle ABC$---(i)   (Angles opposite to equal sides are equal)

In $\vartriangle ACD$,

$AC=AD$

This implies,

$\angle ADC=\angle ACD$---(ii)   (Angles opposite to equal sides are equal)

In $\vartriangle BCD$,

$ \angle DBC+\angle BCD+\angle BDC=180^o$   (Angle sum property of a triangle)

$\angle ACB+\angle BCD+\angle ACD=180^o$   (From equations i and ii) 

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

$\angle BCD+\angle BCD=180^o$

$2(\angle BCD)=180^o$

$\angle BCD=\frac{180^o}{2}$

$\angle BCD=90^o$

The measure of $\angle BCD$ is $90^o$.

Hence proved.

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Updated on: 10-Oct-2022

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