Ex.10.2 Q13 Circles Solution - NCERT Maths Class 10
Question
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.