# Perpendicular bisector that pass through a fixed point

Points A and B fixed, and point C moves on circle such that ABC acute triangle. $$AT = BT$$ and $$TM \perp AC, \, TN \perp BC$$. How can I proove that all the middle perpendiculars (perpendicular bissector) to $$MN$$ passes through a fixed point?

• And what is variable in your problem? – Aqua Sep 21 '19 at 12:34
• Well if everything is fixed then every point on a line is fixed point. – Aqua Sep 21 '19 at 12:51
• Josef, do you mean “such that ABC is always acute”? – Anton Sherwood Sep 21 '19 at 21:03

Hint:Let's mark perpendicular bisector as (PB). If C is coincident on A or B then M and N will be coincident on A and B respectively and the (PB) of MN is exactly the (PB) of AB.Also when the triangle is isosceles MN is parallel with AB, so again (PB) of MN and AB are coincident. That is the point is on the (PB) of AB. To find this point extend the (PB) of MN to cross the the (PB) of AB at P. It can be shown that the ratio of $$R=\frac{TP}{AB}$$ is independent of position of C and is constant. The angle $$(\alpha)$$ between PT and (PB) of MN is always equal to the angle $$(\beta)$$ between AB and MN(or their extensions),because their rays are perpendicular. If $$\angle CAB$$ or $$\angle CBA$$ is $$90^o$$ Then M or N locate on A or B respectively. Let's mark the intersection of (PB) of MN and AB as Q. The right triangles ABC and PQT are similar . Since AB is constant then TP must be constant due to sine law.That is the equality of angles $$(\alpha)$$ and $$(\beta)$$ requires that (PB) of MN always passes the point P.
Let the tangent at $$A$$ and the tangent at $$B$$ of the circumscribing circle $$\mathcal{C}$$ of $$\triangle ABC$$ meet at $$U$$ and let $$V$$ be the midpoint of $$TU$$. We assert that $$V$$ is the required fixed point through which the perpendicular bisector of $$MN$$ passes as $$C$$ varies over $$\mathcal{C}$$.
First we claim that $$CU \perp MN$$. This is equivalent to proving that $$\angle NCT=\angle UCM$$. To this end, we must verify that $$CP$$ bisects $$\angle TCU$$ where $$P$$ is the midpoint of the arc $$AB$$ of $$\mathcal{C}$$ not containing $$C$$ (note also that $$CP$$ bisects $$\angle ACB$$). This means we have to show that $$\mathcal{C}$$ is the circle of Apollonius of the segment $$TU$$. In other words, if $$Q$$ is the point such that line segment $$PQ$$ is a diameter of $$\mathcal{C}$$ we must show that $$\frac{|PU|}{|PT|}=\frac{|QU|}{|QT|}.$$
Now if $$r$$ is the radius of $$\mathcal{C}$$ and $$h$$ is the length $$|TB|$$, then it follows easily that $$|OT|=\sqrt{r^2-h^2}$$ and $$|OU|=\frac{r^2}{\sqrt{r^2-h^2}}$$ where $$O$$ is the center of $$\mathcal{C}$$. So $$|PU|=\frac{r^2}{\sqrt{r^2-h^2}}-r,$$ $$|PT|=r-\sqrt{r^2-h^2},$$ $$|QU|=\frac{r^2}{\sqrt{r^2-h^2}}+r,$$ $$|QT|=r+\sqrt{r^2-h^2}.$$ That is $$\frac{|PU|}{|PT|}=\frac{r}{\sqrt{r^2-h^2}}=\frac{|QU|}{|QT|},$$ finishing the claim. (An alternative proof is to notice that $$\angle PBT=90^\circ-\angle OTB = 90^\circ-\left(90^\circ-\frac12\angle TOB\right)=\frac12\angle TOB$$ and $$\angle UBP=\angle UBO-\angle OBP=90^\circ-\left(90^\circ-\frac12\angle TOB\right)=\frac12\angle TOB,$$ so $$PB$$ bisects $$\angle UBT$$ internally. Because $$Q$$ is on the line $$UT$$ and $$\angle PBQ=90^\circ$$, $$QB$$ bisects $$\angle UBT$$ externally.)
Now the dilation $$d$$ about $$T$$ with factor $$+1/2$$ sends the straight line $$CU$$ to a line $$\ell$$ perpendicular to the segment $$MN$$. Since $$\ell$$ passes through the midpoint of $$CT$$, which is the center of the circumscribing circle of the cyclic quadrilateral $$CMTN$$, $$\ell$$ must be a perpendicular bisector of $$MN$$. Because $$d(U)=V$$, $$\ell$$ passes through $$V$$ as asserted.