1. **State the problem:** Given two triangles $DEF$ and $GHI$ with right angles at $E$ and $H$ respectively, and the proportion $\frac{DE}{GH} = \frac{EF}{HI}$ along with $xE \cong xH$, prove that $\frac{EF}{HI} = \frac{DF}{GI}$. 2. **Recall the properties:** In right triangles, the Pythagorean theorem applies: $$DF^2 = DE^2 + EF^2$$ and $$GI^2 = GH^2 + HI^2$$. Also, congruent angles imply similarity or certain equalities in corresponding sides. 3. **Use the given proportion:** $$\frac{DE}{GH} = \frac{EF}{HI} = k$$ for some constant $k$. This means $$DE = k \cdot GH$$ and $$EF = k \cdot HI$$. 4. **Express $DF$ and $GI$ using Pythagoras:** $$DF = \sqrt{DE^2 + EF^2} = \sqrt{(k GH)^2 + (k HI)^2} = \sqrt{k^2 GH^2 + k^2 HI^2} = k \sqrt{GH^2 + HI^2} = k \cdot GI$$. 5. **Form the ratio:** $$\frac{DF}{GI} = \frac{k \cdot GI}{GI} = k$$. 6. **Recall from step 3:** $$k = \frac{EF}{HI}$$. 7. **Therefore:** $$\frac{EF}{HI} = \frac{DF}{GI}$$, which is what we needed to prove. **Final answer:** $$\boxed{\frac{EF}{HI} = \frac{DF}{GI}}$$