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. $\endgroup$ Feb 14, 2020 at 5:18
  • 1
    $\begingroup$ What you are trying to prove is not true $\endgroup$ Feb 14, 2020 at 6:15

1 Answer 1


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, 2020 at 5:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.