# Prove that the circumcenter of $\triangle ABT$, $\triangle EFT$ and the midpoint of $KT$ collinear. $$MO \cap (O) = E, F$$. $$MAB$$ and $$MN$$ are respectively a secant and a tangent of $$(O)$$ at $$N$$ ($$ME < MA < MN < MB < MF$$).The tangent of $$(O)$$ at $$E$$ intersects semicircle diameter $$MF$$ at $$K$$. $$NO \cap FK = T$$. Prove that the circumcentre of $$\triangle ABT$$, $$\triangle EFT$$ and the midpoint of $$KT$$ collinear.

I am trying to prove the following claims leading to the solution of the problem.

• $$KN \perp MT$$

• $$MT \perp PQ$$ (or $$MT$$ is the tangent of two externally circumcircles of $$\triangle ABT$$ and $$\triangle EFT$$)

• $$RS \parallel KN$$ (or $$S$$ is the midpoint of $$NT$$ with $$S = NT \cap PQ$$, $$P$$ and $$Q$$ are respectively the circumcentre of $$\triangle ABT$$ and $$\triangle EFT$$)

This is supposed to be an easy problem. Why couldn't I solve it?

Consider three circles: the circumcircle of $$ATB$$, the circumcircle of $$ETF$$ and the circle with diameter $$KT$$. All of them pass through $$T$$. To show that their centers lie on a line, we will show that they have common radical axis. $$T$$ has the same power $$0$$ with respect to the circles.
Now, let's prove that M has the same power with repect to the circles. Firstly, $$MA\cdot MB=ME\cdot MF$$. Secondly, the triangle $$MKF$$ is right, and $$KE$$ is its height to hypotenuse, so $$ME\cdot MF=MK^2$$