Asked by | 18th Feb, 2010, 07:11: AM
Let ABCD be a quadrilateral circumscribing a circle with centre O.
Join AO, BO, CO, DO.
The angles BAO and DAO formed at A are equal, as AB and AD are tangents.
Let each of these angles equal a.
The angles at B are similarly equal to each other. Let each of them equal b.
Similarly for vertices C and D.
The sum of the angles at the centre is 360 deg.
The sum of the angles of ABCD is 360 deg.
2(a + b + c + d) = 360
a + b + c + d = 180.
From triangle AOB, angle BOA = 180 - (a + b).
From triangle COD, angle COD = 180 - (c + d).
Angle BOA + angle COD = 360 - (a + b + c + d)
= 360 - 180
= 180 deg.
Thus AB and CD subtend supplementary angles at O.
Answered by | 18th Feb, 2010, 11:40: AM
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