Question

In the given figure, $\angle 1=\angle 2$ and $\frac{AC}{BD}=\frac{CB}{CE}$, prove that $\Delta$ACB$\sim \Delta$DCE.

Answer 1

We have, $\frac{AC}{BD}=\frac{CB}{CE}$

$⟹\frac{AC}{CB}=\frac{BD}{CE}$

$⟹\frac{AC}{CB}=\frac{CD}{CE}(\because ,BD=DCas\angle 1=\angle 2)$........(i)

Also, $\angle 1=\angle 2$
i.e. $\angle DBC=\angle ACB$
$\therefore \Delta ACB\sim \Delta DCE$ ( by SAS similarity criterion)
Hence proved.