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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.
Solution
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
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