In Fig. 4.145, if AB ⊥ BC, DC ⊥ BC and DE ⊥ AC, prove that ∆CED ∼ ∆ABC.

It is given that $A B \perp B C, D C \perp B C$ and $D E \perp A C$.

We have to prove that $\triangle C E D \sim \triangle A B C$.

Now,

$A B \perp B C, D C \perp B C$, so $A B \| D C$.

In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{CED}$,

$\angle B=\angle E=90^{\circ}$ (Given)

$\angle \mathrm{A}=\angle \mathrm{ECD}$ ( Alternate angles)

So, $\triangle C E D \sim \triangle A B C$ (AA similarly rule)