Question:
In Fig. 4.145, if $\mathrm{AB} \perp \mathrm{BC}, \mathrm{DC} \perp \mathrm{BC}$ and $\mathrm{DE} \perp \mathrm{AC}$, prove that $\triangle \mathrm{CED} \sim \triangle \mathrm{ABC}$.
Solution:
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)