Q5 of 56 Page 119

In figure, if DE || BC and CD || EF then prove that AD2 = AB × AF.

In ΔADE & ΔABC,



∠A = ∠A |common angle


∠ADE = ∠ABC |corresponding angles


ΔADE~ ΔABC by AA Similarity Rule


…(1)


In ΔAFE & ΔADC,



∠A = ∠A |common angle


∠AFE = ∠ADC |corresponding angles


ΔAFE~ ΔADC by AA Similarity Rule


…(2)


From (1) & (2),



⇒ AD2 = AB × AF


Hence, proved.


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