Question

AB and CD are parallel straight lines. EG intersects the lines AB and CD at F and G respectively. If $\angle AFE={60}^{\circ }$ and the straight line HI bisects $\angle EGD$, then $\angle DGI$ is:

• A. ${60}^{\circ }$
• B. ${120}^{\circ }$
• C. ${130}^{\circ }$
• D. ${140}^{\circ }$

Answer 1

The correct option is B ${120}^{\circ }$

$AB$ $\parallel$ $CD$ ....given

$\angle EFA={60}^o$ ...given

$\therefore \angle EGC={60}^o$ ....corresponding angles

Now, $\angle EGC$ and $\angle EGD$ form a linear pair.

$\therefore \angle EGC+\angle EGD={180}^o$ ....(1)

But $HI$ is the bisector of $\angle EGD$

$\therefore \angle EGD=2\angle HGD$ ....(2)

Substitute eq (2) in eq(1), we get

$\angle EGC+2\angle HGD={180}^o$

$\therefore \angle HGD=\frac{{180}^o-{60}^o}{2}={60}^o$ ....(3)

Now, $\angle HGD$ and $\angle DGI$ form a linear pair.

$\therefore \angle DGI={120}^o$.