Question

In figure, $\angle BAC={90}^o$, AD is its bisector. If DE $\perp$ AC, Prove that $DE\times (AB+AC)=AB\times AC$.

Answer 1

It is given that AD is the bisector of $\angle Aof△ABC$.
$\therefore \frac{AB}{AC}=\frac{BD}{DC}$
$\Rightarrow \frac{AB}{AC}+1=\frac{BD}{DC}+1$ [Adding 1 on both sides]
$\Rightarrow \frac{AB+AC}{AC}=\frac{BD+DC}{DC}$
$\Rightarrow \frac{AB+AC}{AC}=\frac{BD}{DC}$ ...(i)
In ${△}^{\prime }s$ CDE and CBA, we have
$\angle DCE=\angle BCA=\angle C$ [Common]
$\angle BAC=\angle DEC$ [Each equal to ${90}^o$]
$△CDE\sim △CBA$
Corresponding sides will be in proportion
so $\frac{CD}{CB}=\frac{DE}{BA}=\frac{CE}{CA}$
$\frac{AB+AC}{AC}=\frac{BC}{DC}=\frac{AB}{DE}$
we get $DE\times (AB+AC)=AB\times AC$.