Question

In the given figure, BC⊥AB,AE⊥ABBC⊥AB,AE⊥AB. Shoe that $DE\cdot BC=AD\cdot AB$

Answer 1

We have

$\angle ABC=\angle ADE$ [Each equal to ${90}^{\circ }$]

$\angle ACB=\angle EAD$ [Alternate angles]

$\therefore$ By $AA$ similarity

$\Delta ABC\sim \Delta EDA$

$\Rightarrow \frac{BC}{AB}=\frac{AD}{DE}$

$\Rightarrow DE\cdot BC=AD\cdot AB$. $\left[henceproved\right]$