Question

Prove, $DE(AB+AC)=AB\times AC$

Answer 1

In $△AED$
Let us consider $AE=x$
$AE=DE$ (Isosceles triangle with $\angle EAD=\angle ADE={45}^{\circ }$)
$\therefore AE=DE=x$
$AED$ is an Right angled triangle.
$\therefore A{E}^2+D{E}^2=A{D}^2$
$x^2+x^2=A{D}^2$
$2x^2=A{D}^2$
$\sqrt{2x^2}=AD$
$\sqrt{2}x=AD$
In $△AED$ & $△CED$
$\angle ADE=\angle CDE$ (each ${45}^{\circ }$)
$DE=DE$ (Common side)
$\angle AED=\angle CED$ (each ${90}^{\circ }$)