IN TRIANGLE ABC,DE||BC AND CD||EF PROVE THAT AD SQUARE= AF Sq×AB
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Here is the proof. Sorry for the scratches!
Hope it helps!
Hope it helps!
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Given : DE || BC and CD || EF
Prove that : AD 2 = AB x AF
Proof : DE || BC (Given (ΔABC) )
AB/AD = AC/AE (Basic Proportionality Theorem)
FE || DC (Given (ΔADC)
AD/AF = AC/AE (Basic Proportionality Theorem )
AB/AD = AD/AF ( From (2) and (4) (Transitivity)
AD2 = AB x AF (Cross multiply_)
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