prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.....♡
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Step-by-step explanation:
∆A0P Congruent to ∆A0S[As AP=AC,OP=OS,AO=AO Common]
Angle 1=Angle 8
2=3
4=5
6=7 ....... (i)
==>1,2,3,4,5,6,7,8=360°
...==>2Angle1+2Angle2+2Angle6+2Angle5=360°
..==>(Angle1+Angle2)+(Angle6+Angle 5)
==>Angle AOB+Angle COD=180°
Similarly,
Angle AOD+Angle BOC=180°.