Question
If two lines intersect, prove that the vertically opposite angles are equal.
✓
Answer
Given, two lines $AB$ and $CD$ intersect at point $O$.
To prove
$(i)\ce{ \angle AOC =\angle BOD}$
$(ii)\ce{ \angle AOD=\angle BOC}$
Proof:
$(i)$ Ray $OA$ stands on line $CD.$
$\therefore\ce{ \angle AOC + \angle AOD}=180^{\circ} [$linear pair axiom$] ...(i)$ Ray $OD$ stands on line $AB$.
$\therefore \ce{\angle AOD + \angle BOD}=180^{\circ} [$linear pair exiom $] \ldots (ii)$
From the equations $(i)$ and $(ii),$
$\Rightarrow \angle \ce{AOC + \angle AOD=\angle AOD + \angle BOD}$
$\ce{\angle AOC=\angle BOD}$
$(ii)$ Ray $OD$ stands on line $AB$.
$ \therefore \ce{\angle AOD + \angle BOD}=180^{\circ}[$ linear pair axiom $] \ldots (iii)$
Ray $OB$ stands on line $CD.$
$ \therefore \ce{\angle DOB + \angle BOC=180^{\circ}}$
From equations $(iii)$ and $(iv)$
$\Rightarrow \ce{\angle AOD + \angle BOD=\angle DOB+\angle BOC}$
$\ce{\angle AOD=\angle BOC}$
Hence proved.
To prove
$(i)\ce{ \angle AOC =\angle BOD}$
$(ii)\ce{ \angle AOD=\angle BOC}$
Proof:
$(i)$ Ray $OA$ stands on line $CD.$
$\therefore\ce{ \angle AOC + \angle AOD}=180^{\circ} [$linear pair axiom$] ...(i)$ Ray $OD$ stands on line $AB$.
$\therefore \ce{\angle AOD + \angle BOD}=180^{\circ} [$linear pair exiom $] \ldots (ii)$
From the equations $(i)$ and $(ii),$
$\Rightarrow \angle \ce{AOC + \angle AOD=\angle AOD + \angle BOD}$
$\ce{\angle AOC=\angle BOD}$
$(ii)$ Ray $OD$ stands on line $AB$.
$ \therefore \ce{\angle AOD + \angle BOD}=180^{\circ}[$ linear pair axiom $] \ldots (iii)$
Ray $OB$ stands on line $CD.$
$ \therefore \ce{\angle DOB + \angle BOC=180^{\circ}}$
From equations $(iii)$ and $(iv)$
$\Rightarrow \ce{\angle AOD + \angle BOD=\angle DOB+\angle BOC}$
$\ce{\angle AOD=\angle BOC}$
Hence proved.
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