# Prove that opposi

Let circle with centre O touches the sides AB, BC, CD and DA of a quadrilateral ABCD at the points P, Q, R and S respectively. Then we have to prove that AOB + COD = 180° and AOD + BOC = 180°

Now, join OP, OQ, OR and OS.

Since the two tangents drawn from an external point to a circle subtend equal angles at the centre,

1 = 2, 3 = 4, 5 = 6 and 7 = 8 … (1)

Now,

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 360° … (2)

[Since sum of all the angles subtended at a point is 360°]

From (1) and (2),

2 (2 + 3 + 6 + 7) = 360°

(2 + 3) + (6 + 7) = 180°

AOB + COD = 180°

Similarly,

2 (1 + 8 + 4 + 5) = 360°

(1 + 8) + (4 + 5) = 180°

AOD + BOC = 180°

Hence proved.

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