parmenides51 wrote: As shown in figure , in the pentagon $ABCDE$, $BC = DE$, $CD \parallel BE$, $AB>AE$. If $\angle BAC = \angle DAE$ and $\frac{AB}{BD}=\frac{AE}{ED}$. Prove that $AC$ bisects the line segment $BE$. It’s suffice to show that $ABDE$ is cyclic quadrilateral because then it is harmonic quadrilateral and so $AD$ is $A$ symmedian of triangle $ABE$, implying that $AC$ bisects $BE$. (Btw We already have $B,C,D,E$ concylic.) But I guess the hard part of this problem is proving the concylic.