In an isosceles ∆ABC, the equal sides BA and CA are produced through A upto E and F respectively such that AE = AF. If K and L are the mid points of FB and EC, prove that FB = EC and AK = AL.
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Answer:
To prove : AB×EF=AD×EC
⇒ECAB=EFAD
Proof :
AB=AC (∵ ABC is isosceles)
∴∠B=∠C (angles opposite to equal sides are equal) - (1)
In ΔABD and ΔECF
∠ABD=∠ECF (from (1))
∠ADB=∠EFC (Both are 90∘)
Using AA similarity
ΔADB∼ΔECF
⇒ECAB=EFAD
⇒AB×EF=AD×EC
∴ Hence proved.
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