Show this quad is cyclic

OP mentions this blog post as the inspiration for the question. That post might be saying that $AIEF'$ is cyclic because of "radical axis", but if so it makes the error of assuming that $I,D,F'$ are collinear.

The blog post refers to a thread on the problem, of which this is one entry. The entry defines $F'$ as collinear with $I,D$, and in a later step (after showing $AIEF'$ cyclic) demonstrates that $F'$ is collinear with $E,M'$.

Assuming $I,D,F'$ collinear the "radical axis" method amounts to computing powers of the point $D$ with respect to the circles $(ABC)$ and $(BIC)$ and the three lines $AE,BC,IF'$, giving us $$AD\cdot DE=BD\cdot DC=ID\cdot DF'.$$ By the intersecting chords theorem $AD\cdot DE=ID\cdot DF'$ implies $AIEF'$ is cyclic.

After this, we can angle chase (as in the thread entry) to show that $F'$ is collinear with $E,M'$ giving us the harmonic quad setup.