Q32 of 186 Page 263

In the given figure, ∠BAC = 90° and AD ⊥ BC. Then,

∠ABD + ∠ BAD = 90°


∠ ABD + (90-∠CAD) = 90°
 ∠ ABD = ∠ DAC
In ΔBDA and Δ ADC,
∠ ABD = ∠ CAD
∠ BDA = ∠ ADC = 90°
Therefore, ΔBDA and Δ ADC are similar by AAA.
BD.CD = AD2
Therefore the correct option is (c).

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