Question

The straight lines $AB$ and $CD$ intersect at $E$. If $EF$ and $EG$ are bisectors of $\angle DEA$ and $\angle AEC$
respectively and if $\angle AEF=x$ and $\angle AEG=y$, then

• A. $x+y>{90}^o$
• B. $x+y<{90}^o$
• C. $x+y={90}^o$
• D. $x+y={180}^o$

Answer 1

The correct option is C $x+y={90}^o$

$EF$ is the bisector of $\angle AED$

$\therefore \angle AED=\angle FED=x$

$EG$ is the bisector of $\angle AEC$

$\angle AEG=\angle GEC=y$

Now $\angle AEG+\angle GEC+\angle AED+\angle FED={180}^{\circ }$ (Angles made on straight line)

$y+y+x+x={180}^{\circ }$

$2x+2y={180}^{\circ }$

$x+y={90}^{\circ }$