Question

in the adjoining fig prove that
1.) AC||FD
2.)BF||EC​

Answer 1

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