# A circle touches

The diagram is as follows: Let PQRS be the quadrilateral which has a circle inside it with center O.

Now join AO, BO, CO, DO

From the figure, 1 = 8 [Two tangents drawn from an external point to a circle subtend equal angles at the centre.]

We can also state that 2 = 3, 4 = 5 and 6 = 7

Sum of angles in a quadrilateral is 360°

Sum of angles at the center is 360°

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 360°

2(1 + 2 + 5 + 6) = 360°

1 + 2 + 5 + 6 = 180°

or

2(8 + 3 + 4 + 7) = 360°

8 + 3 + 4 + 7 = 180°

POQ + ROS= 180°

Also QOR + POS = 180°

[Since POQ = 1 + 2

ROS = 5 + 6

QOR = 7 + 8

POS = 3 + 4

as shown in the figure]

Hence it is proved that PQ and RS subtend supplementary angles at center.

Thus the angles subtended at the centre by a pair of opposite sides are supplementary.

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