Question

In isosceles triangle triangle $ABC$, $AB=AC$. The side $BA$ is produced to $D$ such that $BA=AD$. Prove that : $\angle BCD={90}^{\circ }$.

Answer 1

Const: Join CD.

In triangle $ABC$
$AB=AC$
$\angle ACB$=$\angle ABC$......(i)
In triangle $ACD$
$AC=AD$
$\angle ADC$=$\angle ACD$.......(ii)
Adding (i) and (ii)

$\angle ABC+\angle ADC$ $=\angle ACB+\angle ACD$
$\therefore \angle ABC+\angle ADC=\angle BCD$.......(iii)
[from figure, $\angle ACB+\angle ACD=\angle BCD$]

In triangle $BCD$, we have
$\angle ABC+$$\angle ADC$$+\angle BCD=180$
$\angle BCD+\angle BCD=180$ [From equation(iii)]
$2\angle BCD=180$
$\therefore \angle BCD=90$