In the figure,$OD$ is the bisector of $\angle AOC$,$OE$ is the bisector of $\angle BOC$,and $OD \perp OE$. Show that the points $A, O$,and $B$ are collinear.

(N/A) Given: In the figure,$OD \perp OE$. $OD$ and $OE$ are the bisectors of $\angle AOC$ and $\angle BOC$ respectively.
To prove: Points $A, O$,and $B$ are collinear,i.e.,$AOB$ is a straight line.
Proof: Since $OD$ and $OE$ bisect $\angle AOC$ and $\angle BOC$ respectively,
$\angle AOC = 2 \angle DOC$ ---$(1)$
$\angle BOC = 2 \angle COE$ ---$(2)$
Adding equations $(1)$ and $(2)$,we get:
$\angle AOC + \angle BOC = 2 \angle DOC + 2 \angle COE$
$\Rightarrow \angle AOC + \angle BOC = 2(\angle DOC + \angle COE)$
$\Rightarrow \angle AOC + \angle BOC = 2 \angle DOE$
Since $OD \perp OE$,$\angle DOE = 90^{\circ}$.
$\Rightarrow \angle AOC + \angle BOC = 2 \times 90^{\circ} = 180^{\circ}$.
Since the sum of adjacent angles $\angle AOC$ and $\angle BOC$ is $180^{\circ}$,they form a linear pair. Therefore,$AOB$ is a straight line,and points $A, O$,and $B$ are collinear.