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

 A: 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.


*

*All angles marked with the same color are equal.

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

*H is the orthocenter of the isosceles triangle DBF.

*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.
A: I found a solution using only Euclid geometry knowledge, but it's not in 9th-level, so I still hope to see some pure primitive solution.

To make the problem easier, I reverse the question to be like this: Let Q be the intersection of the line through M which is parallel with AB and BC. Then we just need to prove that N, Q and I are in a line
Let K is the intersection of BC and MN. BC is the polar line of A with (O), so we must have (MNKD) = -1. Therefore, (IN,IM,IK,ID) = -1. Let P is the intersection of MQ and NI and L is the intersection of MQ and IK. Because MP is parallel with ID, we must have L is midpoint of MP. 
Using the triangle KBD, because I is midpoint of BD and MQ is parallel to BD, we have L is midpoint of MQ, too. So we conclude that P and Q coincide.
A: I'm lucky to find this beautiful solution, and it only uses 9th-grade level knowledge to solve. Here is the solution
Let P and R are intersections of BC with OI and AM, correspondingly. Let S is the reflection of M through I.
Firstly, even using 9th-grade level, we can easily prove that ${RM \over RN} = {AM \over AN}$, therefore ${RM \over MA} = {RN \over AN}$. I is the same midpoint of both BA and MS, hence BSAM is a parallelogram, therefore BS is parallel to MR. 
We have this equation chain:

$${PR \over PB} = {RM \over BS} = {RM \over MA} = {NR \over NA}$$

Therefore, we must have that PN is parallel to BA. Therefore, we have this equation chain:

$${PQ \over QB} = {PN \over BI} = {PN \over IA} = {MN \over MA}$$

Therefore, QM is parallel to BA. The problem is solved.
