Question

In the given figure, $\angle ABD=\angle BDC$ and $CD=4AB$. Show that $BD=5BE$.

Answer 1

Given : $\angle ABD=\angle BDC$ and $CD=4AB$

Since $E$ is common in both the triangles,

$\therefore$ $\angle AEB=\angle CED$

Also, $\angle EAB=\angle ECD$ ........... (each ${90}^o$)

$\therefore$ $△ABE\sim △CDE$

Hence,

$\frac{AB}{BE}=\frac{CD}{DE}$

$⟹$ $\frac{AB}{BE}=\frac{4AB}{DE}$

$⟹$ $DE=4BE$ .................. $(CD=4AB)$

$\therefore$ $DB=BE+ED=BE+4BE=5BE$.

$\therefore$ $BD=5BE$.