Question

In the figure, if $AB\perp BC$, $DC\perp BC$ and $DE\perp AC$, prove that $△CED$ $\sim △ABC.$

Answer 1

$\angle B=\angle C=\angle E={90}^o$ [ Given ]

$\Rightarrow$ $\angle B+\angle C={90}^o+{90}^o={180}^o$

$\therefore$ $AB\parallel DC$

When $AB\parallel DC$ cuts by transversal $AC$

$\angle BAC=\angle ACD$

$\angle BAC=\angle ECD$

$\angle B=\angle E={90}^o$ (given) ...(ii)

$\angle ACB=\angle EDC$ (due to (i) and (ii))

$△ABC\sim △CED$ [By AAA similarity]