Question

If two lines intersect, prove that the vertically opposite angles are equal.

Answer 1

Solve to prove the given statement.

Given: Let us assume the two pairs of vertically opposite angles to be,

$1$^st pair $–\angle AOC$ and$\angle BOD$

$2$^nd pair $–\angle AOD$ and $\angle BOC$

To prove:

We have to prove that vertically opposite angles are equal,

i.e., $\angle AOC=\angle BOD$

$\angle AOD=\angle BOC$

From the diagram above

The ray $AO$stands on the line $CD.$

We know that,

$\Rightarrow \angle AOC+\angle AOD=180°$ (By linear pair axiom) $...\left(i\right)$

$\Rightarrow \angle AOD+\angle BOD=180°$ (By linear pair axiom) $...\left(ii\right)$

From equations $\left(i\right)$ and $\left(ii\right)$,

We obtain

$\angle AOC+\angle AOD=\angle AOD+\angle BOD$

$\Rightarrow$$\angle AOC=\angle BOD...\left(iii\right)$

$\Rightarrow \angle DOB+\angle COB=180°$ (By linear pair axiom) $...\left(iv\right)$

Also, the ray $CO$ lies on the line $AOB.$

$\Rightarrow \angle COB+\angle AOC=180°$(By linear pair axiom) $...\left(v\right)$

From equations $\left(iv\right)$ and $\left(v\right)$,

We get,

$\angle DOB+\angle COB=\angle COB+\angle AOC$

$\Rightarrow$ $\angle DOB=\angle AOC...\left(vi\right)$

Thus, from equation$\left(iii\right)$ and equation $\left(vi\right),$

We have,

$\angle AOC=\angle BOD,$ and $\angle DOB=\angle AOC$

Thus, we get, vertically opposite angles are equal.

Hence Proved.