Let ABC be a triangle. Outside the triangle on its edges we build rhombi CBKL

Question

Let ABC be a triangle. Outside the triangle on its edges we build rhombi CBKL, ACMN, BAOP so that K lies on the line AB, M lies on the line BC, and O lies on the line CA. Prove that the lines AL, BN, CP intersect at a common point.

I'm trying to apply Ceva's Theorem to solve this question, but haven't been able to make any progress.

Answer 1

Let $AL\cap BC=\{A_1\}$,$BN\cap AC=\{B_1\}$ and $CP\cap AB=\{C_1\}.$

Thus, $ABLC$ is a trapezoid and in the standard notation we obtain: $$\frac{BA_1}{A_1C}=\frac{AB}{CL}=\frac{AB}{BC}=\frac{c}{a}.$$ Similarly, $$\frac{CB_1}{B_1A}=\frac{a}{b}$$ and $$\frac{AC_1}{C_1B}=\frac{b}{c}.$$ Id est, $$\frac{BA_1}{A_1C}\cdot\frac{CB_1}{B_1A}\frac{AC_1}{C_1B}=1$$ and we are done by the Ceva's theorem for $\Delta ABC$.