Question

$ABC$ is a right angled triangle such that $AB=AC$ and bisector of $\angle C$ intersects the side $AB$ at $D$, then prove that $AC+AD=BC$.

Answer 1

Let $AB=AC=a$ and $AD=b$.

$△ABC$ is right angled at A $\Rightarrow$ $A{B}^2+A{C}^2=B{C}^2$ $\Rightarrow BC=a\sqrt{2}$

$BD=AB-AD=a-b$

By angle-bisector theorem, $\frac{AD}{BD}=\frac{AC}{BC}$.

$\Rightarrow \frac{b}{a-b}=\frac{a}{a\sqrt{2}}=\frac{1}{\sqrt{2}}$

$\Rightarrow b=\frac{a}{1+\sqrt{2}}=a(\sqrt{2}-1)$

$\Rightarrow a+b=a\sqrt{2}$

But, we know $AC=a$, $AD=b$, $BC=a\sqrt{2}$

$\Rightarrow AC+AD=BC$. Hence it is proved.