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Question 13
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.


Solution


Let ABCD be a quadrilateral circumscribing a circle with O such that it touches the circle at point P, Q, R, S. Join the vertices of the quadrilateral ABCD to the center of the circle.
In Δ OAP and Δ OAS,
AP = AS (Tangents from the same point)
OP = OS (Radii of the circle)
OA = OA (Common side)
Δ OAP Δ OAS (SSS congruence condition)
POA=AOS
1=8
Similarly we get,
2=3
4=5
6=7
Adding all these angles,
1+2+3+4+5+6+7+8=360
(1+8)+(2+3)+(4+5)+(6+7)=360
21+22+25+26=360
2(1+2)+2(5+6)=360
(1+2)+(5+6)=180AOB+COD=180
Similarly, we can prove that BOC+DOA=180
Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Mathematics

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