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Question
A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC ?
Solution
Let the sides of the quadrilateral ABCD touch the circle at points P, Q, R and S as shown in the figure.
We know that, tangents drawn from an external point to the circle are equal in length.
Therefore,
`{:[AP,=,AS],[BP,=,BQ],[CQ,=,CR],[DR,=,DS]}` ................(1)
∴AB + CD = (AP + BP) + (CR + DR)
= (AS + BQ) + (CQ + DS) [Using (1)]
= (AS + DS) + (BQ + CQ)
= AD + BC
Hence, AB + CD = AD + BC
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