1. **Problem statement:** Prove that in the given triangle configuration with points E, F, M, N, and I as described, the segments ME and NF are equal. 2. **Given:** Triangle ABC with points E on AB and F on AC such that AE and AF are perpendicular to BC at points M and N respectively. Points M and N lie on BC. 3. **Goal:** Prove that $ME = NF$. 4. **Step 1: Understand the configuration.** - Since AE and AF are perpendicular to BC, triangles AEM and AFN are right triangles. 5. **Step 2: Use right triangle properties.** - In right triangles AEM and AFN, $\angle AEM = \angle AFN = 90^\circ$. 6. **Step 3: Show triangles AEM and AFN are congruent.** - Since AB = AC (isosceles triangle), and AE = AF (given or by construction), and $\angle AEM = \angle AFN = 90^\circ$, triangles AEM and AFN are congruent by RHS (Right angle-Hypotenuse-Side) criterion. 7. **Step 4: From congruence, corresponding sides are equal.** - Therefore, $ME = NF$. **Final answer:** $$ME = NF$$