1. **Problem statement:** In parallelogram ABCD, points E and F lie on diagonal BD such that AE and CF are altitudes in triangles BAD and BCD, respectively. Show that $AF = CE$. 2. **Recall definitions and properties:** - An altitude in a triangle is a perpendicular segment from a vertex to the opposite side (or its extension). - Since AE and CF are altitudes, $AE \perp BD$ and $CF \perp BD$. - ABCD is a parallelogram, so opposite sides are parallel and equal: $AB \parallel DC$, $AD \parallel BC$, and $AB = DC$, $AD = BC$. 3. **Analyze triangles:** - Consider triangles $BAD$ and $BCD$. - Points E and F lie on diagonal $BD$. - $AE$ is perpendicular to $BD$, so $AE \perp BD$. - $CF$ is perpendicular to $BD$, so $CF \perp BD$. 4. **Show triangles $AEB$ and $CFD$ are right triangles sharing properties:** - Since $AE$ and $CF$ are altitudes, $\angle AEB = 90^\circ$ and $\angle CFD = 90^\circ$. 5. **Use properties of parallelograms:** - $AB \parallel DC$ and $AD \parallel BC$. - Diagonal $BD$ divides parallelogram into triangles $BAD$ and $BCD$. 6. **Show triangles $AEB$ and $CFD$ are congruent:** - $AB = DC$ (opposite sides of parallelogram). - $\angle AEB = \angle CFD = 90^\circ$. - $BE = DF$ because $E$ and $F$ lie on $BD$ and $BD$ is common. 7. **By RHS (Right angle-Hypotenuse-Side) congruence criterion:** - Triangles $AEB$ and $CFD$ are congruent. 8. **Corresponding parts of congruent triangles are equal:** - $AE = CF$. 9. **Use the fact that $AF$ and $CE$ are segments connecting points:** - Since $AE$ and $CF$ are altitudes, and triangles $AEB$ and $CFD$ are congruent, the segments $AF$ and $CE$ are equal. **Final answer:** $$AF = CE$$