in the adjoining fig prove that
1.) AC||FD
2.)BF||EC
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Step-by-step explanation:
given in the picture
AB = ED
Angle ABC = Angle DEF
AC = FE
According to S-A-S
∆ ABC ~= ∆DEF
then FE = BC
BE = BE
then FB = EC
now
FC divides the Parallel lines AC and FD
angle ACB= Angle DEF (CPCT)
so AC||FD