Find the flaw in the proof of showing diagonals of a parallelogram bisect each other.

Question

There is something wrong with the proof of the following theorem. Find the error and correct the proof.

Theorem: The diagonals of a parallelogram bisect each other.

Proof: Let $ABCD$ be a parallelogram with diagonals $AC$ and $BD$ intersecting at $O$. $AD$ is parallel to $BC$ and $AB$ is parallel to $DC$ since $ABCD$ is given as a parallelogram. Since alternate interior angles formed by parallel lines are congruent, angle $CAD$ is congruent so angle $ACB$ and angle $ADB$ is congruent to angle $DBC$. $AD$ is congruent to $BC$ since opposite sides of a parallelogram are equal in length. Thus triangle $AOD$ is congruent to triangle $COB$ by ASA congruence. Thus, $AO$ is congruent to $OC$ and $BO$ is congruent to $OD$ since corresponding parts of congruent triangles are congruent.

I am supposed to find a flaw in this proof, but to me it appears flawless.

Answer 1

The proof states that opposite sides of a parallelogram are congruent. This is of course true but not axiomatic. So one who gives the proof shown may be asked to justify the lemma that opposite sides are in fact congruent. (Exercise for the reader: Do this.)

Incidentally, if you are given a quadrilateral where opposite sides are congruent, you can prove the diagonals bisect each other without reference to parallel lines or Euclidean geometry, e.g. it works on a sphere. You first draw one diagonal and prove the resulting triangles congruent by SSS, so then each of two opposing vertex angles is divided the same way. Apply that result to a quadrilatetal with both diagonals drawn and prove opposite triangles congruent by ASA as above. Stating this formally is another thing the reader can do.