  Question

Prove that opposite sides of a quadrilatral circumscribing a ciecle subtend supplementary angle at the centre of the circle. Solution

Given: Let ABCD be the quadrilateral circumscribing the circle with centre O.ABCD touches the circle at points P, Q, R and S.To prove: Opposite sides subtends supplementary angles at centre.i.e., $$\angle AOB+\angle COD=180^o$$& $$\angle AOD+\angle BOC=180^o$$Construction: Join OP, OQ, OR and OS.Proof: Let us rename the anglesIn $$\Delta AOP$$ and $$\Delta AOS$$,AP$$=$$AS(Length of tangents drawn from external point to a circle are equal)AO$$=$$AO (common)OP$$=$$OS (both radius)$$\Delta AOP\cong \Delta AOS$$(SSS congruence rule)$$\angle AOP=\angle AOS$$(CPCT)i.e., $$\angle 1=\angle 8$$ …………….$$(1)$$Similarly, we can prove$$\angle 2=\angle 3$$ ……………….$$(2)$$$$\angle 5=\angle 4$$ ………………..$$(3)$$$$\angle 6=\angle 7$$ ……………….$$(4)$$Now,$$\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8=360^o$$(sum of angles round a point is $$360^o$$)$$\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 1=360^o$$$$\angle 1+\angle 2+\angle 5+\angle 6=\dfrac{360^o}{2}$$$$(\angle 1+\angle 2)+(\angle 5+\angle 6)=180^o$$$$\angle AOB+\angle COD=180^o$$Hence  both angles are supplementary.similarly, we can prove$$\angle BOC+\angle AOD=180^o$$Hence proved. Mathematics

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