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

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.

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

• What is outercircle? – Aqua Jul 29 '18 at 11:14
• What is a "fixed line"? – Jens Jul 29 '18 at 11:15
• Sure the "outer circle" refers to the circumcircle of the triangle. A "fixed line" refers to a line that doesn't depend on the choice of the point $M$ on the line $Ax$, etc. – John Hughes Jul 29 '18 at 11:17
• But then $AB$ is obviously – Aqua Jul 29 '18 at 11:18
• It's certainly obvious that $AB$ contains a point of the circumcircle (namely $O$). It's not obvious to me that $AB$ is the tangent to the circumcircle at $O$, independent of the choice of $M$, but I'm not that good at geometry, so it may be obvious to others. – John Hughes Jul 29 '18 at 11:25

## 2 Answers

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 $AB$.
• Great ............+1 – Aqua Jul 29 '18 at 11:37
• Why $OK$ is parallel to $MA$ ? – Word Shallow Jul 29 '18 at 12:22
• @WordShallow By the converse of intercept theorem. – Intelligenti pauca Jul 29 '18 at 13:28

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, $AO=HO=BO=R$.

• How do you know M,N,H are colinear? – Aqua Jul 29 '18 at 11:37
• Prove $MN$ is tangent with half circle at $H$. – Word Shallow Jul 29 '18 at 11:52
• How do you find that? – Aqua Jul 29 '18 at 11:52