Question

In triangle ABC if BD perpendicular to AC prove that (AB-BC)(AB+BC)=(AD+DC)(AD-DC)

Answer 1

In ∆BDA,

Since <BDA =90°,

By pythagorus therm,

BA²=BD²+DA²............(1)

In ∆BDC,

Since <BDC =90°,

By pythagorus therm,

BC²=BD²+DC²............(2)

The given eqn,

(AB-BC)(AB+BC)=(AD+DC)(AD-DC)

Since (a+b)(a-b)= a² - b²

Therefore:

AB²-BC²=AD²-DC²

LHS=

AB²-BC²

(BD²+DA²) - (BD²+ DC²)

BD²+DA²-BD²-DC²

DA²-DC²

RHS