Question

ABCD is a cyclic quadrilateral .Prove that AC.BD = AB.DC + AD.BC​

Answer 1

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Answer 2

AC.BD = AB.DC + AD.BC​     Proved

Step-by-step explanation:

Consider the provided information.

Let the length of AB =a, BC = b, CD = c, DA = d

According to the Law of Cosines:  $c^2 = a^2 + b^2 - 2ab \cos C$

In ΔABC,

$AC^2 = a^2 + b^2 - 2ab \cos B$

and ΔADC,

$AC^2 = c^2 + d^2 - 2cd \cos D$

and B + D = π

⇒ cos B + cos D = 0

⇒ $AC^2 (cd + ab) = (a^2 + b^2) cd + (c^2+ d^2) ab$

⇒ $AC^2 = (a^2cd+ c^2ab) (b^2cd +d^2ab)cd+ ab$        ...(1)

Similarly by taking another diagonal BD,

$BD^2=(ba+ cd)(bd+ ca)da +bc$  ...(2)

Multiplying equation (1) and (2)

⇒ $(AD \cdot BD)^2 = (ac + bd)^2$

⇒ $AC \cdot BD = ac + bd$

⇒ $AC \cdot BD = AB \cdot CD + BC\cdot AD$

Hence, proved

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