Let's begin by marking the parallel segments in the given diagram and the congruent angles.
Notice that EFGH is a parallelogram, and therefore ∠ E ≅ ∠ G. In consequence, by the Angle-Angle (AA) Similarity Postulate we conclude that △ EFK ~ △ GFJ.
Finally, by definition of similar triangles, we obtain the required relation.
△ EFK ~ △ GFJ
⇒
EK/KF = GJ/JF ✓
Two-Column Proof Table
In the table below we summarize the proof we did before.
