(N/A) Given: Two lines $AB$ and $CD$ intersect at point $O$.
To prove:
$(i)$ $\angle AOC = \angle BOD$
$(ii)$ $\angle AOD = \angle BOC$
Proof:
$(i)$ Since ray $OA$ stands on line $CD$,by the linear pair axiom:
$\angle AOC + \angle AOD = 180^{\circ}$ ....$(1)$
Similarly,since ray $OD$ stands on line $AB$,by the linear pair axiom:
$\angle AOD + \angle BOD = 180^{\circ}$ ....$(2)$
From equations $(1)$ and $(2)$,we get:
$\angle AOC + \angle AOD = \angle AOD + \angle BOD$
Subtracting $\angle AOD$ from both sides,we get:
$\angle AOC = \angle BOD$
Hence,proved.
$(ii)$ Since ray $OD$ stands on line $AB$,by the linear pair axiom:
$\angle AOD + \angle BOD = 180^{\circ}$ ....$(3)$
Similarly,since ray $OB$ stands on line $CD$,by the linear pair axiom:
$\angle BOD + \angle BOC = 180^{\circ}$ ....$(4)$
From equations $(3)$ and $(4)$,we get:
$\angle AOD + \angle BOD = \angle BOD + \angle BOC$
Subtracting $\angle BOD$ from both sides,we get:
$\angle AOD = \angle BOC$
Hence,proved.