Triangles — Maths STD 10 — Question

Given: In the figure $\text{AB}\perp\text{BC}, \text{DC} \perp\text{BC}$ and $ \text{DE} \perp \text{AC}$
To prove: $\triangle\text{CED}\sim\triangle\text{ABC}$.
Proof: $\text{AB}\perp \text{BC}$
$\angle\text{B} = 90^\circ$
and $\angle\text{A}+\angle\text{ACB}=90^\circ\ ...(\text{i})$
$\text{DC}\perp \text{BC}$
$\angle\text{DCB}=90^\circ$
$\Rightarrow\angle\text{ACB} + \angle\text{DCA} = 90^\circ ….(\text{ii})$
From (i) and (ii)
$\angle\text{A} = \angle\text{DCA}$
Now in $\triangle\text{CED}$ and $\triangle\text{ABC,}$
$\angle\text{E} = \angle\text{B}$ (each 90°)
$\angle\text{DEA or } \angle\text{DCE} = \angle\text{A}$ (proved)
$\triangle\text{CED} \sim \triangle\text{ABC}$ (AA axiom)
Hence proved.