On the side $AD$ of the convex quadrilateral $ABCD$ with an acute angle at $B$, a point $E$ is marked. It is known that $\angle CAD = \angle ADC=\angle ABE =\angle DBE$. (Grade 9 version) Prove that $BE+CE<AD$. (Grade 10 version) Prove that $\triangle BCE$ is isosceles.(Here the condition that $\angle B$ is acute is not necessary.)