In the two-column proof with given $\overline{FG} \parallel \overline{HI}$, what reason completes step 4 to prove $\triangle EFG \sim \triangle EHI$?

What the proof is trying to finish

You already established two angle congruences, one shared angle at $E$ and one pair created by the parallel lines. The last line needs a similarity theorem that uses angle information.

Match the angles to the triangles

• Step 2: $\angle E \cong \angle E$ means the triangles share the angle at vertex $E$.
• Step 3: $\angle G \cong \angle I$ comes from corresponding angles when a transversal intersects the parallel lines $\overline{FG}$ and $\overline{HI}$.

So the triangles $\triangle EFG$ and $\triangle EHI$ have two pairs of corresponding congruent angles.

Why this is enough to claim similarity

If two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar. This is exactly the AA Similarity Postulate.

Therefore, step 4 is justified by AA Similarity, giving $\triangle EFG \sim \triangle EHI$.