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.

  • $\begingroup$ 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. $\endgroup$ – math wizard Feb 14 '20 at 5:18
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    $\begingroup$ What you are trying to prove is not true $\endgroup$ – 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.

  • $\begingroup$ Can you wpload a picture? $\endgroup$ – Mick Feb 15 '20 at 5:33

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