Question

In a triangle ABC; the angle BAC = 90° and AD is drawn perpendicular to BC. Prove that
AD^2 = BD.DC.​

Answer 1

Given : In a triangle ABC; the angle BAC = 90° and AD is drawn perpendicular to BC.

Solution:

ΔDAC      and ΔDBA

∠DAC =  ∠B     as both are 90° - C

∠ADC  = ∠BDA  = 90°

=> ΔDAC  ~   ΔDBA

=> AD/ BD  =  DC/ AD

=> AD² = BD . DC

QED

Hence Proved

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