Let us draw the figure.

Here, we can see that

AB and CD intersect each other at point E.

The two pairs of vertically opposite angles are -

1^st pair - ∠AEC and ∠BED

2^nd pair - ∠AED and ∠BEC

We need to prove that the vertically opposite angles are equal, i.e.,

∠AEC = ∠BED, and ∠AED = ∠BEC

Now, we can see that the ray AE stands on the line CD. We know, if a ray stands on a line then the sum of the adjacent angles is equal to 180°.

⇒ ∠AEC + ∠AED = 180° (By linear pair axiom) - - - - (i)

Similarly, the ray DE stands on the line AEB.

⇒ ∠AED + ∠BED = 180° (By linear pair axiom) - - - - (ii)

From equations (i) and (ii), we have

∠AEC + ∠AED = ∠AED + ∠BED

⇒ ∠AEC = ∠BED - - - - (iii)

Similarly, the ray BE stands on the line CED.

⇒ ∠DEB + ∠CEB = 180° (By linear pair axiom) - - - - (iv)

Also, the ray CE stands on the line AEB.

⇒ ∠CEB + ∠AEC = 180° (By linear pair axiom) - - - - (v)

From equations (iv) and (v), we have

∠DEB + ∠CEB = ∠CEB + ∠AEC

⇒ ∠DEB = ∠AEC - - - - (vi)

Thus, from equation (iii) and equation (vi), we have

∠AEC = ∠BED, and ∠DEB = ∠AEC

Therefore, it is proved that the vertically opposite angles are equal.