Prove parallelogram when opposite angle pair congruent and one diagonal is intersected in the middle by the other

Question

In the convex quadrilateral ABCD, AC intersects BD at the point {O}.
It is known that AO is congruent with OC and angle ABC is congruent with angle ADC.

Prove that ABCD is a parallelogram.

Notes:

• the problem is from a 7th grade book and is supposed to be solved by a 13 years old kid using Euclidean knowledge
• I tried to prove that the opposite triangles formed by the diagonals are congruent, using the middle point theorem, but wasn't able
• I also thought that Homothety properties or Desargues theorem may help, but I would not be able to explain this to the kid.

Answer 1

I don't know if that technique is familiar to a 7th grader, but the theorem can be proved by RAA.

Suppose by contradiction $OD\not\cong OB$ and assume WLOG that $OD<OB$. Then we can construct $E$ inside $OB$ such that $OD\cong OE$.

Then $AECD$ is a parallelogram and $$ \angle AEC\cong\angle ADC\cong \angle ABC. $$

But that is not possible, because $\angle AEC > \angle ABC$ by Euclid's exterior angle theorem.