Question

In given figure, $\angle BAC={90}^0$, 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 $\angle A$ of $△ABC$.

$\therefore$ $\frac{AB}{AC}=\frac{BC}{DC}$

$\Rightarrow$ $\frac{AB}{AC}+1=\frac{BD}{DC}+1$ [Adding 1 on both sides]

$\Rightarrow$ $\frac{AB+AC}{AC}=\frac{BD+DC}{DC}$

$\Rightarrow$ $\frac{AB+AC}{AC}=\frac{BC}{DC}$........(i)

In ${△}^{\prime }sCDEandCBA$, we have

$\angle DCE=\angle BCA=\angle C$ [Common]

$\angle BAC=\angle DEC$ [Each equal to ${90}^{\circ }$]

so, by AA-criterion of similarity, we have
$△CDE\sim △CBA$

$\Rightarrow$ $\frac{CD}{CB}=\frac{DE}{BA}$

$\Rightarrow$ $\frac{AB}{DE}=\frac{BC}{DC}$ ........(ii)

From (i) and (ii), we have

$\frac{AB+AC}{AC}=\frac{AB}{DE}$

$\Rightarrow$ $DE\times (AB+AC)=AB\times AC$ $\left[Henceproved\right]$