The two triangles have a pair of corresponding sides as well as the
included angles between them that are equal.
Justin could prove that ΔDEH ≅ ΔFGH by A. SAS
Reason:
The known information are;
DF bisects GE, therefore; GH = EH
GE bisects DF, therefore; DH = FH
∠DHE = ∠FHG by vertical opposite angle theorem
Therefore;
DH and EH in triangle ΔDEH = FH and GH in triangle ΔFGH
∠DHE in triangle ΔDEH = ∠FHG in triangle ΔFGH
Therefore, two sides and an included angle triangle in ΔDEH are equal to (congruent) to the corresponding two sides and an included angle in ΔFGH.
Therefore;
ΔDEH ≅ ΔFGH by SAS, Side-Angle-Side congruency theorem
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