Question

In isosceles triangle ABC, AB = AC. The side BA is produced to D such that BA = AD.
Is $\angle BCD={90}^{\circ }$. ?
If true then enter 1 else if False enter 0

Answer 1

In triangle ABC,
AB = AC, so , $\angle ABC=\angle ACB$
Now, since, AB =AD then, AC = AD
hence, $\angle ACD=\angle ADC$
Now, let $\angle ABC=\angle ACB=x$ and $\angle ACD=\angle ADC=y$
so, $\angle ABC+\angle BCD+\angle BDC=180$
$\angle ABC+\angle ACB+\angle ACD+\angle BDC=180$
$x+x+y+y=180$
$2(x+y)=180$
$x+y=90$
$\angle ACB+\angle ACD=90$
$\angle BCD=90$