Ex.10.2 Q13 Circles Solution - NCERT Maths Class 10

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.

Text Solution

To prove:

Opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.

Steps and Reasoning:

We know that, tangents drawn from an external point to a circle subtend equal angles at the center.

In the above figure, \({P,\;Q,\;R,\;S}\) are point of contacts

\({AS = AP}\) (The lengths of tangents drawn from an external point \(A\) to a circle are equal.)

\[\begin{align} \angle {SOA} = \angle {POA} = \angle 1 = \angle 2 \end{align}\]

Tangents drawn from an external point to a circle subtend equal angles at the centre.

Similarly,

\[\begin{align} \angle 3 = \angle 4 , \angle 5 = \angle 6 , \angle 7 = \angle 8 \end{align}\]

Since complete angle is \({360^ \circ }\),

\[\begin{align} \angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8 &=360^{\circ} \\ 2(\angle 1+\angle 8+\angle 4+\angle 5) &=360^{\circ} \\ \angle 1+\angle 8+\angle 4+\angle 5 &=180^{\circ} \end{align}\]

[or]

\[\begin{align} 2(\angle 2+\angle 3+\angle 6+\angle 7) &=360^{\circ} \\ \angle 2+\angle 3+\angle 6+\angle 7 &=180^{\circ} \\ \angle A O B+\angle C O D &=180^{\circ} \end{align}\]

From above figure \(\begin{align}\angle A O D+\angle B O C =180^{\circ}\end{align}\)

\(\angle {AOD}\) and \(\angle {BOC}\) are angles subtended by opposite sides of quadrilateral circumscribing a circle and sum of them is \({180^ \circ }\)

Hence proved.