Question

Two triangles $\Delta ABC$ and $\Delta DBC$ are on the same base $BC$ and on the same side of $BC$ in which $\angle A=\angle D={90}^{}.$ If $CA$ and $BD$ meet each other at $E,$ then show that $AE\times EC=BE\times ED.$

Answer 1

Given triangles $ABC$ and $DBC$ are n the same base $BC$.

Consider ${\Delta }^{\prime }s$ $ABC$ and $DBC$

$\angle A=\angle D={90}^0$ (given)

$\angle AEB=\angle DEC$ (vertically opposite angles are equal )

Hence,

$\Delta ABC\sim \Delta DBC$ (AA similarity theorem)

$\begin{array}{c}\Rightarrow \frac{AE}{DE}=\frac{BE}{CE}\\ \therefore AE\times EC=BE\times ED\end{array}$

Hence, proved.