Question

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

Answer 1

Step-by-step explanation:

Let AB = a, BC = b, CD = c, DA = d

using cosine rule in ΔABC and ΔADC, we get :

AC2 = a2 + b2 – 2ab cos B

AC2 = c2 + d2 – 2cd cos D

and B + D = π

⇒ cos B + cos D = 0

⇒ AC2 (cd + ab) = (a2 + b2) cd + (c2+ d2) ab

⇒ AC2 = (a2cd+ c2ab) (b2cd +d2ab)cd+ ab

Similarly by taking another diagonal BD, we can show that :

BD2=(ba+ cd)(bd+ ca)da +bc

Multiplying the two equations

⇒ (AD . BD)2 = (ac + bd)2

⇒ AC . BD = ac + bd

⇒ AC . BD = AB . CD + BC . AD

Answer 2

Answer:

Answer in attachment.

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