The following is a lemma someone has presented to me:
Lemma:
Let $ABCD$ be a parallelogram in the plane, and let $O$ be a point outside of the parallelogram. Prove $\angle OBA = \angle ADO$ if and only $\angle BOC = \angle AOD$, where all angles are directed.
For example, here is a diagram: 
However, my attempts to find the proof and name, if one exists, of this lemma have not been successful. I have tried my hand at proving this myself, but I’m not all the best at geometry, so I didn’t make much progress.
Here’s a sketch of what I did:
I considered the isogonal conjugation of the points $A$ and $C$ with respect to triangle $OBD$. If we assume $\angle OBA = \angle ADO$, then both conjugates lie on the perpendicular bisector of $BC$, and if $\angle BOC = \angle AOD$, then the isogonal conjugate of $A$, $A^*$, lies on $OC$, and vice versa. However this method falls apart as it doesn’t encode the fact that $AB \parallel CD$ in a neat way.
Any hints or input will be appreciated.
Let $X$ be the midpoint of $AC$. Let $A', C'$ be the reflections of $A$ and $C$ with respect to $OD, OB$, respectively. Let $Y, Z$ be the midpoints of $CC', AA'$, respectively. Note that $Y, Z$ are the perpendicular projections of $C, A$ onto $OB, OD$, repsectively.
You can prove that $\angle BOC = \angle AOD \iff \triangle AOC' \equiv \triangle A'OC \iff AC' = A'C \iff XY=XZ.$
Now, construct a parallelogram $AOCO'$ and note that $\angle ODA = \angle O'BC$. This is equivalent to $\angle OBC = \angle ABO'$. Let $T$ be the projection of $A$ onto $O'B$. Repeat the argument above to show the equivalence $\angle OBC = \angle ABO' \iff XY = XT$. Combine all ingredients keeping in mind that $XT=XZ$ due to symmetry.