If two lines intersect at a point, then the vertically opposite angles are always equal.

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We know that the sum of angles lying on a straight line is ${{180}^{\circ }}$.Let us consider the angles on the line CD. So, we get $\angle COB+\angle BOD={{180}^{\circ }}$.$\Rightarrow \angle COB={{180}^{\circ }}-\angle BOD$ ---(1).Now, let us consider the angles on the line AB. So, we get $\angle COB+\angle AOC={{180}^{\circ }}$ ---(2).Let us substitute equation (2) in equation (1).So, we get ${{180}^{\circ }}-\angle BOD+\angle AOC={{180}^{\circ }}$.$\Rightarrow \angle AOC={{180}^{\circ }}-{{180}^{\circ }}+\angle BOD$.$\Rightarrow \angle AOC=\angle BOD$.From the figure, we can see that the angles $\angle AOC$ and $\angle BOD$ are vertically opposite angles.So, we have proved that if two lines intersect each other, then the vertically opposite angles are equal.

Note: We can also prove that the other pair of vertically opposite angles $\angle COB$ and $\angle DOA$ equal in the similar way as shown below:

Let us consider the angles on the line CD. So, we get $\angle COB+\angle BOD={{180}^{\circ }}$.$\Rightarrow \angle BOD={{180}^{\circ }}-\angle COB$ ---(3).Now, let us consider the angles on the line AB. So, we get $\angle BOD+\angle DOA={{180}^{\circ }}$ ---(4).Let us substitute equation (3) in equation (4).So, we get ${{180}^{\circ }}-\angle COB+\angle DOA={{180}^{\circ }}$.$\Rightarrow \angle DOA={{180}^{\circ }}-{{180}^{\circ }}+\angle COB$.$\Rightarrow \angle DOA=\angle COB$.We use this result to get the required answers.

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If two lines intersect at a point, then the vertically opposite angles are always equal.

When two lines intersect each other, then the opposite angles, formed due to intersection are called vertical angles or vertically opposite angles

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