Question

In the given figure, $\angle 1=\angle 2$ and $\angle 3=\angle 4$. Show that $AE\times BC=AC\times DE$.

Answer 1

In $△$ADE and $△$ABC,

As $\angle 1=\angle 2$

$\Rightarrow \angle 1+\angle BAE=\angle 2+\angle BAE$

OR

$\angle DAE=\angle BAC$

$\angle 3\angle 4$

$\therefore$ $△$ ADE\sim $△$ABC (AA similarity)

$\frac{AD}{AB}=$ $\frac{AE}{AC}$ $\frac{DE}{BC}$

$\therefore$ AE.BC=AC.DE