Question

ABC is a right triangle with angle B = 90º. A circle with BC as diameter meets by hypotenuse AC at point D. Prove that: BD² = AD × DC.

Answer 1

In ΔADB,

∠D = 90°

∴ ∠A + ∠ABD = 90° ...(i)

But in ΔABC, ∠B = 90°

∴ ∠A + ∠C = 90°   ...(ii)

From (i) and (ii)

∠C = ∠ABD

Now in ΔABD and ΔCBD

∠BDA = ∠BDA = 90°

∠ABD = ∠BCD

∴ ΔABD ∼ ΔCBD  ...(AA postulate)

∴ `(BD)/(DC) = (AD)/(BD)`

`=>` BD² = AD × DC