In the figure AABC, BELAC and CFLAB. Then which of the following a) AE.EC=AF.AC
b) AE.FC=AF.EB
c) AB.BC=AC.EB
d) AE.BC=AB.CF
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Given AB=AC
∴∠B=∠C (1)
In △ABD and △ECF
∠B=∠C (From 1)
∠ADB=∠EFC=90
o
[∵AD⊥BC and EF⊥AC,]
∴ By AA Criterion of Similarity, △ABD ∼ △ECF hence proved
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