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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