Question

In given figure, DEFG is a square and $\angle BAC={90}^0$. Prove that

(i) $△AGF\sim △DBG$

(ii) $△AGF\sim △EFC$

(iii) $△DBG\sim △EFC$

(iv) ${DE}^2=BD\times EC$

Answer 1

(i) In $△AGF$ and $△DBG$, we have

$\angle GAF=\angle BDG$ [EAch equal to ${90}^0$] and,

$\angle AGF=\angle DBG$ [Corresponding angles]

$\therefore$ $△AGF\sim △DBG$ [By AA-criterion of similarity] $\left[Henceproved\right]$

(ii) In $△AGF$ and $△EFC$, we have

$\angle FAG=\angle CEF$ [Each equal to ${90}^0$] and,

$\angle AFG=\angle ECF$ [Corresponding angles]

$\therefore$ $△AGF\sim △EFC$ [By AA-criterion of similarity] $\left[Henceproved\right]$

(iii) Since $△AGF\sim △DBG$ and $△AGF\sim △EFC$

$\therefore$ $△DBG\sim △EFC$ $\left[Henceproved\right]$

(iv) we have,

$△DBG\sim △EFC$ [Using (iii)]

$\therefore$ $\frac{BD}{EF}=\frac{DG}{EC}$

$\Rightarrow$ $\frac{BD}{DE}=\frac{DE}{EC}$ [$\because$ DEFG is a square $\therefore$ EF=DE, DG=DE]

$\Rightarrow$ ${DE}^2=BD\times EC$ $\left[Henceproved\right]$