1. **State the problem:** Prove that $\angle E \cong \angle H$ given the similarity of triangles $\triangle DEF \sim \triangle GHI$ and the ratio $\frac{DE}{GH} = \frac{DF}{GI} = EF$. 2. **Recall the similarity rule:** If two triangles are similar by SSS similarity, then their corresponding angles are congruent and their corresponding sides are proportional. 3. **Given:** $$\frac{DE}{GH} = \frac{DF}{GI} = EF$$ and $$\triangle DEF \sim \triangle GHI$$ 4. **By SSS similarity, corresponding angles are congruent:** $$\angle E \cong \angle H$$ 5. **Explanation:** Since the triangles are similar by SSS, all corresponding sides are in proportion, and all corresponding angles are congruent. Therefore, the angle at vertex $E$ in $\triangle DEF$ corresponds to the angle at vertex $H$ in $\triangle GHI$, so $\angle E \cong \angle H$. **Final answer:** $$\boxed{\angle E \cong \angle H}$$