Prove that in the given pentagon ABCDE AB||CE In a convex pentagon $ABCDE$ we have  $BC\parallel AD$, $CD\parallel BE$, $DE\parallel AC$ and $AE\parallel BD$.
Prove that $AB\parallel CE$.
 A: 
Consider the above figure. All the angles of the same color are equal , using properties of parallel lines. Also , using the fact that the opposite sides of parellelograms are equal , all the line segments of the same color are equal. We are required to prove that $\bf{AB\parallel CE}$
First , we join $M_1J_1$ , $N_1J_1$ , $N_1K_1$ and $M_1L_1$. Using the converse of Thales’ Theorem , we can prove that they are parallel to segments $EA\&N_1L_1$,$BC \& M_1K_1$, $AC$ and $BE$ respectively. Denote the blue segments by $b$ , the yellow segments by $y$ , $J_1L_1$ by $x$ and $K_1J_1$ by $z$ . Applying Thales’ Theorem to the required $\triangle$s , we have $$\frac{EM_1}{M_1C}=\frac{b}{x+b}=\frac{x}{b}=\frac{M_1N_1}{N_1C}$$ $$\frac{CN_1}{N_1E}=\frac{y}{z+y}=\frac{z}{y}=\frac{N_1M_1}{EM_1}$$
From these two equations , we get $EM_1=N_1C$ And from this , we get $$\frac{x}{b}=\frac{z}{y}$$ Therefore , $$\frac{J_1E}{J_1B}=\frac{J_1C}{J_1A}$$ Therefore , $\triangle J_1AB \sim \triangle J_1CE $ by $SAS$ similarity. 
It follows that $\bf{CE \parallel AB}$
