Question

In figure,ray OC stands on the line AB; ray OL and ray OM are angle bisector of $\angle AOC$ and $\angle BOC$, respectively.Prove that $\angle LOM$ = 90${}^{\circ }$ [3 MARKS]

Answer 1

Concept: 1 Mark
Application: 2 Marks

In figure, we have

$\angle AOC+\angle BOC={180}^{\circ }$ (Linear pair of angles) ... (1)

Now, ray OL bisects $\angle AOC$,

$\angle LOC=\frac{1}{2}\angle AOC$ .....(2)

$\angle MOC=\frac{1}{2}\angle BOC$ ..... (3)

Adding (2) and (3), we have

$\angle LOC+\angle MOC=\frac{1}{2}\angle AOC+\frac{1}{2}\angle BOC$

$\angle LOM=\frac{1}{2}\angle AOC+\frac{1}{2}\angle BOC$

$\angle LOM=\frac{1}{2}\times {180}^{\circ }={90}^{\circ }$

$\angle LOM={90}^{\circ }$

Hence Proved