A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC.

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Solution

We know that the tangents drawn from an external point to a circle are equal.

$∴ A P = A S \dots \dots (i) [tangents from A] B P = B Q \dots \dots \dots . (i i) [tangents from B] C R = C Q \dots \dots \dots \dots . (i i i) [tangents from C] D R = D S \dots \dots \dots . . (i v) [tangents from D] ∴ A B + C D = (A P + B P) + (C R + D R) = (A S + B Q) + (C Q + D S) [using (i), (i i), (i i i) and (i v)] = (A S + D S) + (B Q + C Q) = (A D + B C) Hence, (A B + C D) = (A D + B C)$

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