In ∆ ABC, m angle B= 90° . seg AD is perpendicular to side BC. If AC= 20,AB=15 , BC= 12 ; than BD = ?
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Since , ∆ABC is right triangle (right angled at A) , so by Pythagoras theorem , we say :
AB² + AC² = BC²….(1)
Also AD is perpendicular to BC which implies that ∆ BDA and ∆ CDA are also right triangles : so we again apply Pythagoras theorem to get following equations :
AD² + DC² = AC²….(2)
BD² + AD² = AB²….(3)
Adding equations (2) and (3) :
BD² + DC² + 2AD² = AC² + AB²
From (1) we replace RHS of above equation to get :
BD² + DC² + 2AD² = BC²
But BC = (BD + CD)
So we again replace :
BD² + DC² + 2AD² = (BD + DC)²
Expanding we get :
BD² + DC² + 2AD² = BD² + CD² + 2BD×CD
Cancelling and simplifying we get :
AD² = BD × CD
Hence, proved.
However it is a long method and not as effecient as the other two answers are by using similarity and trigonometry.
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