We are told that GH ∥ FJ, and from the diagram we see that FH is a transversal. Then, by the Alternate Interior Angles Theorem we get ∠ JFH ≅ ∠ GHF.
Notice that FH is a common side for both △ HJF and △ FGH, and by the Reflexive Property of Congruent Segments we have FH ≅ FH. Besides, ∠ J ≅ ∠ G because they have the same measure.
cc
∠ J ≅ ∠ G & Angle
∠ JFH ≅ ∠ GHF & Angle
FH ≅ FH & Non-included Side
Finally, by the Angle-Angle-Side (AAS) Congruence Postulate we get △ HFJ ≅ △ FGH. We summarize the proof in the following two-column table.
