Question

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

Answer 1

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

In a right angled triangle $ABC,{BC}^2={AB}^2+{AC}^2$

$\Rightarrow {BC}^2=a^2+a^2=2a^2$

$\Rightarrow BC=a\sqrt{2}$

Given $AD=b,$ we get

$DB=AB-AD=a-b$

We have to prove that $AC+AD=BC$ or $a+b=a\sqrt{2}$

By the angle bisector theorem, we get

$\frac{AD}{DB}=\frac{AC}{BC}$

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

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

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

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

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

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

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

or $AD+AC=BC$

Hence proved.