Question

plzzzzzz very urgent

Answer 1

in ∆ABC and ∆DEF,
i)edf=acb
ii)efd =cba(alternate angles)

bd=cf
or, bd+cd=cf+cd
iii)or, bc=df
therefore , ∆abc congruent to ∆def(by a.s.a)
ac=de(cpctc)

Answer 2

In ACB and EDF

<ACB=<EDF( given)

<EFD= <CBA (alt angles since EF II AB and BF is the transversal)

ACB~ EFD

AC= DE ( CPST)