# In convex pentagon $ABCDE$ we have $BC=DE,\angle ABE=\angle CAB=\angle AED-\frac{\pi}{2},\angle ACB=\angle ADE$. Prove that $BCDE$ is parallelogram.

In convex pentagon $ABCDE$ we have $BC=DE$, $\angle ABE=\angle CAB=\angle AED-\frac{\pi}{2}$,$\angle ACB=\angle ADE$. Prove that $BCDE$ is parallelogram.

I proved it by using law of sines in triangles $ABC$ and $ADE$ to determine$\frac{AD}{AC}$ 2 times and when I compared that I obtained $EG=AG$ and then easy by angle chasing. I really want to see elementary proof of this problem.

• Are you familiar with shapes in the Cartesian coordinate system? And dot products? Generically, vectors? Commented Apr 1, 2015 at 17:48
• No and I'm looking for elementary proof Commented Apr 1, 2015 at 18:12
• +1) One start could be to provide us with a picture with the information you gave us... Commented Apr 1, 2015 at 18:13

Reflect line $BE$ about $BA$ and $AE$. Let $X$ be a common point of these two reflections. Then $A$ is incenter of triangle $XEB$, because $BA$ and $EA$ are bisector of angles $EBX$ and $XEB$.

Note that $\angle BAC = \angle EBA = \angle ABX$, so $AC \parallel BX$.

Moreover, $\angle XAE = \frac \pi 2 + \frac {\angle EBX} 2 = \frac \pi 2 + \angle EBA = \angle AED$, so $ED \parallel AX$.

Take $P \in AX$ and $Q \in BX$ such that $APED$ and $ACBQ$ are parallelograms. Let $R$ be reflection of $Q$ about $AX$. Then $R \in EX$.

If $P=Q=R=X$ then we are done. Assume for the sake of contradiction that $X$ does not coincide with any of $P,Q,R$.

Note that $\angle APE = \angle EDA = \angle ACB = \angle BQA = \angle ARE$, therefore points $E,A,P,R$ are concyclic. On the other hand $AP=AQ=AR$.

Clearly $P \neq R$ so $AP=AR$ implies that $\angle REA + \angle PEA = \pi$. We deduce that $\angle PEB = \angle PEA + \angle AEB = \angle PEA + \angle REA = \pi$ which is absurd.