# Proving that a quadrilateral is a parallelogram with the information given.

Let $$ABCD$$ be a quadrilateral. Prove that if $$\overline{AB}$$ is congruent to $$\overline{CD}$$, and $$\angle BCD$$ is congruent to $$\angle DAB$$, then $$ABCD$$ is a parallelogram.

I am feeling stuck because I can't find any avenue to go unless I can somehow prove an angle bisector exists between a diagonal and the pair of congruent angles.

• If I create two diagonals I can show that all four triangles are equal by SSS but then have to prove that the two diagonals meet. – math wizard Feb 14 '20 at 5:18
• What you are trying to prove is not true – Zubin Mukerjee Feb 14 '20 at 6:15

Take $$\Delta ABE$$ such that $$AB=BC$$ and let $$D\in AE.$$
Now, take $$\Delta BCD\cong\Delta DEB$$ such that $$C$$ and $$E$$ are placed in the same half plane respect to $$BD$$.
Thus, $$AB=BE=CD$$ and $$\measuredangle DAB=\measuredangle BED=\measuredangle BCD,$$ but $$ABCD$$ is not parallelogram.