Question

In given figure, AD and BE are respectively perpendiculars to BC and AC. Show that

(i) $△ADC\sim △BEC$

(ii) $CA\times CE=CB\times CD$

(iii) $△ABC\sim △DEC$

(iv) $CD\times AB=CA\times DE$

Answer 1

(i) In ${△}^{\prime }sADC$ and $BEC$, we have

$\angle ADC=\angle BEC={90}^0$ [Given]

$\angle ACD=\angle BCE$ [Common]

So, by AA-criterion of similarity, we have

$△ADC\sim △BEC$ $\left[Henceproved\right]$

(ii) we have,

$△ADC\sim △BEC$ [AS proved above]

$\Rightarrow$ $\frac{AC}{BC}=\frac{DC}{EC}$........(i)

$\Rightarrow$ $CA\times CE=CB\times CD$ $\left[Henceproved\right]$

(iii) In ${△}^{\prime }sABC$ and $DEC$, we have

$\frac{AC}{BC}=\frac{DC}{EC}$ [From(i)]

$\Rightarrow$ $\frac{AC}{DC}=\frac{BC}{EC}$

Also, $\angle ACB=\angle DCE$ [common]

So, by SAS-criterion of similarity, we have

$△ABC\sim △DEC$ $\left[Henceproved\right]$

(iv) we have,

$△ABC\sim △DEC$

$\Rightarrow$ $\frac{AB}{DE}=\frac{AC}{DC}$

$\Rightarrow AB\times DC=AC\times DE$

$\Rightarrow CD\times AB=CA\times DE$ $\left[Henceproved\right]$