$M$ is inside the parallelogram $ABCD$. If $\angle MBA=\angle MDA$, then $\angle MAB = \angle MCB$

Question

Let M be a point inside the parallelogram ABCD, such that $\angle MBA = \angle MDA$ . Prove that $\angle MAB = \angle MCB$.

Source: Romanian Mathematical Gazette no. 5/2023

Answer 1

Draw a line through $M$ parallel to $AD$, $BC$ and draw a point $M'$ such that $MM'=AD=BC$, as on the picture above. This makes the quadrilaterals $AM'MD$ and $M'BCM$ parallelograms.

Thus, $\angle AM'M=\angle ADM=\angle ABM$ which implies that the quadrilateral $AM'BM$ is cyclic.

That in turn implies that $\angle BCM=\angle BM'M=\angle BAM$ (the last equality is implied by $AM'BM$ being cyclic).