Question

solve the question
#anshikarockers​

Answer 1

Step-by-step explanation:

in ∆ ABC,

BD=EC( given )-----1

BD+DE=EC+DE

BE=DC

in ∆BEG and ∆ DFC

angle GEB=angle FDC. ( BA || DF )

BE=DC (from 1 )

angleGBE = angle FDC (CA|| EG)

∆ BEG = ∆ CDF

DG = DF. (C.P.C.T)

EF= CF. ( C.P.C.T)

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