Question

In triangle abc,angle acb=90 and cd perpendicular to ab, prove that cd 2 =bd*ad

Answer 1

To prove CD² = BD × AD
In Δ CAD, CA² = CD² + AD² .... (1)
Also in Δ CDB, CB² = CD² + BD² .... (2)
(1) + (2) we get
CA² + CB² = 2CD² + AD² + BD²
AB² = 2CD² + AD² + BD²
AB² - AD² = BD² + 2CD²
(AB + AD)(AB - AD) - BD² = 2CD²
(AB + AD)BD - BD² = 2CD²
BD(AB + AD - BD) = 2CD²
BD(AD + AD) = 2CD²
BD × 2AD = 2CD²
CD² = BD × AD
Hence proved.

Answer 2

Answer:

Step-by-step explanation:

On triangleABC and triangleACD

∠ACB = ∠CDA,

∠CDA = CAB

ΔABC and ΔACD

 

AC/AB=AD/AC

AC²= AB×AD

similarly ΔBCD and ΔBAC

BC/BA = BD/BC

BC²= BA×BD

∴ ÷ BC² and AC² we get

BC²/AC²=AB×BD/AB×AD

∴ BC²/AC² =BD/AD