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Question

In a cyclic quadrilateral ABCD, if AC=60, prove that the smaller of two is 60.

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Solution

Here, ABCD is a cyclic quadrilateral.


We know, in a cyclic quadrilateral, the sum of opposite angles are supplementary, i.e. 180o.


Then, A+C=180o.


But given, AC=60.

Adding the two equations,
2A=240
A=120.

Therefore, A+C=180o
120o+C=180o
C=60.

The smaller one is 60.

Hence, proved.

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