Step-by-step explanation:
[tex]In \: \triangle DEF \: \& \: \triangle HGF \\ EF \: \cong \: GF...(given) \\ \angle EFD \cong \angle GFH \\ (vertically \: opposite \: angles) \\ \ DF \: \cong \: HF...(given) \\ \therefore \: \triangle DEF \cong \triangle HGF \\ (By\: SAS \:criterion\: of \:congruence) [/tex]
