Question

In triangle ABC ,if angle ACB=90 and CD perpendicular to AB Prove that BC^2/AC^2=BD/AD

Answer 1

Solution :-

To prove CD² = BD × AD

In 

Δ CAD, CA² = CD² + AD² .... (1)

Also in 

Δ CDB, CB² = CD² + BD² .... (2)

(eqⁿ 1) + (eqⁿ 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: