Question

In Fig $3.13$, $\text{line}DE||\text{line}GF\text{ray}EG\text{and ray}FG$ are bisectors of $\angle DEF$ and $\angle DFM$ respectively. Prove that.
(i) $\angle DEG=\frac{1}{2}\angle EDF$
(ii) $EF=FG$

Answer 1

Given : $DE\parallel GF$

$EG$ is bisector of $\angle DEF$

$FG$ is bisector of $\angle DFM$

To proof :

$\left(i\right)$ $\angle DEG=\frac{1}{2}\angle EDF$

$\left(ii\right)$ $EF=FG$

$\because DE\parallel GF$

$\therefore \angle DEF=\angle GFM$ (Corresponding angle)

$2\angle o=\angle x$

$\angle o=\frac{\angle x}{2}⟶\left(1\right)$

$\because 2\angle x=\angle EDF+2\angle o$

$2\times 2\angle o=\angle EDF+2\angle o$

$4\angle o=\angle EDF+2\angle o$

$2\angle o=\angle EDF$

$\angle o=\frac{\angle EDF}{2}$

$\therefore \angle DEG=\frac{\angle EDF}{2}$

In $△DEF$

$\angle DEF+\angle EDF+\angle DFE=180°$

$2\angle o+2\angle o+\angle DFE=180°$

$4\angle o+\angle DFE=180°$

$\therefore \angle DFE=180°-4\angle o$

Now, in $△EGF$

$\angle o+180°-4\angle o+\angle x+\angle EGF=180°$

$\angle o+\angle EGF=3\angle o$ $[\because \angle x=2\angle o]$

$\angle EGF=\angle o$

$\therefore EF=FG$ (side opposite to equal , $\angle o=\angle o$)