Prove that the outer circle of the triangle $OMN$ is always tangent to a fixed line.

Question

Consider a half circle with diameter $AB$.Draw the tangent $Ax$, $By$ with half circle. Take $M$ on $Ax$, $N$ on $By$ such that $AM*BN=R^2$. Prove that the outer circle of the triangle $OMN$ is always tangent to a fixed line.

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I will $MN$ is a fixed line and $O;R$ cut $MN$ at OH is a fixed line

I see: $MN=MA+NB$ and $MN^2=MA^2+NB^2+2MA*NB=MA^2+NB^2+2R^2$

So i So I need to prove $MA^2+NB^2$ is a fixed line but i can't. Help me

Answer 1

HINTS.

1. $MON$ is a right triangle.
2. The circumcircle of $MON$ has its center at $K$, midpoint of $MN$.
3. Radius $OK$ is parallel to $MA$, $NB$ and thus perpendicular to $AB$.

Answer 2

Suppose $HO$ is the perpendicular to $MN$ from $O$.

Notice that

$$ AM \cdot BN = OA\cdot OB = R^2$$ so the triangles $AOM$ and $BON$ are similar.

From here, you obtain that the angle $MON$ is 90, and moreover $MO/MN = AO/NB = BO/NB$ so all the triangles in the figure are similar. In particular, $AO=HO=BO=R$.