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} $. "
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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