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