Prove that the circumcenter of $\triangle ABT$, $\triangle EFT$ and the midpoint of $KT$ collinear.  = 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 $ME\cdot MF=MK^2$

That's all.