Question

In ΔACB, ∠C = 900 and CD ⊥ AB. Prove that BC²/AC² = BD/AD.

Answer 1

From ∆ABC , ∆ACD

<A is common angle ----( 1 )

<ACB = <CDA = 90°-----( 2 )

from ( 1 ) and ( 2 ) ∆ABC ~ ∆ACD

( A.A similarity )

then

BC/CD = AB/AC = AC/AD ,

AB/AC = AC/AD

AB = AC²/AD

AB/AC² = 1/AD

multiply by ' AB ' on both sides

AB/AC² × AB = AB/AD

AB²/AC² = AB/AD

But from ∆ACB ,

AB² = AC² + BC²

[ By Pythagorean theorem ]

(AC²+ BC² )/AC² = AB/AD

AC²/AC² + BC²/AC² = ( AD + BD )/AD

1 + BC²/AC² = AD/AD + BD/AD

1 + BC²/AC² = 1 + BD/AD

Therefore ,

BC²/AC² = BD/AD

Hence proved .

I hope this helps you.

: )