1. **Problem statement:** We have an isosceles triangle ABC with vertex A and base BC. Points M and N lie on BC such that $BM = CN$. Lines perpendicular to BC at M and N intersect AB and AC at points E and F respectively. We need to: a) Show triangles BEM and CFN are congruent. b) Identify homologous elements. c) Deduce the nature of triangle AEF. Then prove triangles CME and BNF are congruent. 2. **Showing triangles BEM and CFN are congruent:** - Since ABC is isosceles with vertex A, $AB = AC$. - Points M and N lie on BC with $BM = CN$. - Lines at M and N are perpendicular to BC, so $\angle BME = \angle CNF = 90^\circ$. - Segments $BM = CN$ by given. - Segments $AB = AC$ by isosceles property. Using the RHS (Right angle-Hypotenuse-Side) congruence criterion: - Right angles at E and F. - Hypotenuses $AB = AC$. - Legs $BM = CN$. Therefore, $\triangle BEM \cong \triangle CFN$. 3. **Homologous elements:** - $B \leftrightarrow C$ - $E \leftrightarrow F$ - $M \leftrightarrow N$ 4. **Nature of triangle AEF:** - Since $\triangle BEM \cong \triangle CFN$, segments $BE = CF$. - Also, $AB = AC$. - Points E and F lie on AB and AC respectively. - By symmetry and congruence, $AE = AF$. Thus, $\triangle AEF$ is isosceles with $AE = AF$. 5. **Proving triangles CME and BNF are congruent:** - Consider triangles CME and BNF. - $CM = BN$ since $BM = CN$ and $BC$ is segment with $B$ and $C$ endpoints. - Angles at E and F are right angles (perpendicular lines). - Segments $CE = BF$ by previous congruence and symmetry. Using RHS criterion again: - Right angles at E and F. - Hypotenuses $CE = BF$. - Legs $CM = BN$. Therefore, $\triangle CME \cong \triangle BNF$. **Final answers:** - $\triangle BEM \cong \triangle CFN$ - Homologous elements: $B \leftrightarrow C$, $E \leftrightarrow F$, $M \leftrightarrow N$ - $\triangle AEF$ is isosceles with $AE = AF$ - $\triangle CME \cong \triangle BNF$