about a ninth-grade geometry problem

My brother asked me this problem, and he is studying ninth-grade. I can't solve it using primitive tools of pure geometry. Hope someone can give me a hint to solve it. Thanks.

Given a circle $(O, R)$ and $A$ is outside $(O)$ such that $OA > 2R$. Draw two tangents AB, AC of $(O)$. Let $I$ is midpoint of AB. Segment OI intersects with (O) at M. AM intersects with (O) at N, $N \neq M$. NI intersects with BC at Q. Prove that MQ perpendicular with OB

Here is the picture

This is NOT a solution. I just want to share some of my findings.

Construction: 1) Extend BO to cut the red circle at D; 2) DA cut the red circle at E and BM extended at F; 3) OE is joined.

By midpoint theorem, we have 1) OMI // DEFA; 2) BJ = JE; BM = MF.

1. All angles marked with the same color are equal.

2. OJMI is the line of centers of the 4 circles and BJHE is the common chord (excluding the green).

3. H is the orthocenter of the isosceles triangle DBF.

4. B, G, H, and M lie on the green circle.

One way to get the Job done is by showing that MQ is parallel to either BI or GH.