Question

in triangle abc
angle c =90 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.