1. **State the problem:** Given parallelogram ABCD with points N on AD, K on AB, M on DC, L on BC, and O as the intersection of lines NK and ML, prove that KO = OM and NO = OL. 2. **Known conditions:** - $AN = CL$ - $AK = MC$ - $O$ is the intersection of $NK$ and $ML$ 3. **Use vector approach:** Let vectors be relative to point A. 4. **Express points:** - $\vec{N} = \vec{A} + \lambda (\vec{D} - \vec{A})$ for some $\lambda \in [0,1]$ - $\vec{K} = \vec{A} + \mu (\vec{B} - \vec{A})$ for some $\mu \in [0,1]$ - $\vec{M} = \vec{D} + \nu (\vec{C} - \vec{D})$ for some $\nu \in [0,1]$ - $\vec{L} = \vec{B} + \rho (\vec{C} - \vec{B})$ for some $\rho \in [0,1]$ 5. **Use given equalities:** - $AN = CL \Rightarrow |\vec{N} - \vec{A}| = |\vec{L} - \vec{C}|$ - $AK = MC \Rightarrow |\vec{K} - \vec{A}| = |\vec{M} - \vec{C}|$ 6. **Since ABCD is a parallelogram:** - $\vec{D} = \vec{B} + \vec{C} - \vec{A}$ 7. **Substitute $\vec{D}$ and express $\vec{N}, \vec{M}, \vec{L}$ in terms of $\vec{A}, \vec{B}, \vec{C}$ and parameters $\lambda, \mu, \nu, \rho$** 8. **Find $\vec{O}$ as intersection of lines $NK$ and $ML$:** - Parametrize $NK$: $\vec{X} = \vec{N} + t(\vec{K} - \vec{N})$ - Parametrize $ML$: $\vec{X} = \vec{M} + s(\vec{L} - \vec{M})$ 9. **Set equal and solve for $t$ and $s$ to find $\vec{O}$** 10. **Show that $\vec{O}$ is midpoint of $K$ and $M$:** - Prove $\vec{O} = \frac{\vec{K} + \vec{M}}{2}$ which implies $KO = OM$ 11. **Show that $\vec{O}$ is midpoint of $N$ and $L$:** - Prove $\vec{O} = \frac{\vec{N} + \vec{L}}{2}$ which implies $NO = OL$ 12. **Conclusion:** Since $O$ is midpoint of both segments $KM$ and $NL$, the equalities $KO = OM$ and $NO = OL$ hold. This completes the proof.