Let $KL$ , the perpendicular bisector of $AB$ cut $AB$, the circumcircle of $~$ $UAB$ at $C$, $M$ $\angle AVM=\angle BVM$ $~$ and since $~$ $UM$ is its diameter, $~$ $\angle UVM=90^{\circ}$ Thus, $~$ $(A,B/N,C)=1$ $~$ and $N$ is the foot of $T$-altitude of $~$ $\triangle TAB$ Here $N$ is the intersection of $VM$ and $AB$

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Let $M$ be the midpoint of $AB$ and $P$ be the intersection of $KL$ and $AB$. $K$ and $L$ are the feet of the altitudes $B$ to $AT$ and $A$ to $BT$, respectively, so $KMNL$ lies on the nine-point circle of $\triangle ABC$. Hence, $(CU)(CV) = (CB)(CA) = (CL)(CK) = (CN)(CM)$, so $MNUV$ is cyclic, so $\angle UVN = 180^{\circ} - \angle UMN = 90^{\circ}$.

Another solution using harmonics. we use the notations of Let $O$ be the midpoint of $AB$, and let $(O)$ be the circle with diameter $AB$. Now, since $T$ lies on the polar of $C$ wrt $(O)$, and since $N$ is the point on $OC$ from such that $TN \perp OC$, therefore, $N$ also lies on the polar of $C$, leading us to $(AB,NC) = -1$. Now, since $O$ is the midpoint of $AB$, therefore, $CN \cdot CO = CA \cdot CB$. But $CA \cdot CB = CU \cdot CV$, therefore, $N,O,U,V$ are concylic. Since $UO \perp AB$, therefore, $UV \perp VN$, and we're done.

Zhero wrote: Let $M$ be the midpoint of $AB$ and $P$ be the intersection of $KL$ and $AB$. $K$ and $L$ are the feet of the altitudes $B$ to $AT$ and $A$ to $BT$, respectively, so $KMNL$ lies on the nine-point circle of $\triangle ABC$. Hence, $(CU)(CV) = (CB)(CA) = (CL)(CK) = (CN)(CM)$, so $MNUV$ is cyclic, so $\angle UVN = 180^{\circ} - \angle UMN = 90^{\circ}$. Do you mean: so $KMNL$ lies on the nine-point circle of $\triangle ABT$. Hence, $(PU)(PV) = (PB)(PA) = (PL)(PK) = (PN)(PM)$ ?

Dear Mathlinkers, (O) is the circle with diameter AB (1) the circle passing through A, U, V, B (2) the Euler's circle of TAB (3) the circle with diameter UN By applying the Monge's theorem (radical axis of three circles are concurrent) we are done. Sincerely Jean-Louis

Under inversion wrt $ C $ with power $ CA.CB $, $ A,B $ interchanges and also $ U,V $ interchanges. Suppose, $ N $ goes to $ N' $. We know inversion preserves cross-ratio. So $ (CN,BA)=(\infty N',AB)=-1 $ So $ N' $ is the mid-point of $ AB $. $ N'U\perp AC $. So $ NV\perp KC $ since $ N'U $ is anti-parallel to $ NV $ wrt $ \angle ACK $. So done.

$\angle KVA=180^{\circ}- \angle AVU=\angle UBA=\angle BAU=\angle BVU$, because quadrilateral $BAVU$ is cyclic. From sine law in triangles $ANC$ and $BNC$ we get: $AN=b\cos \alpha$ and $BN=a\cos \beta$ and $\frac{AN}{BN}=\frac{b \cos \alpha}{a \cos \beta}=\frac{\sin \beta \cos \alpha}{\cos \beta \sin \alpha}$ We also can say that $\frac{AV}{AK}=\frac{\sin \angle AKV}{\sin \angle KVA}$ and $\frac{BV}{BL}=\frac{\sin \angle BLV}{\sin \angle BVL}$, so $\frac{AV}{BV}=\frac{AK\frac{\sin \angle AKV}{\sin \angle KVA}}{BL\frac{\sin \angle BLV}{\sin \angle BVL}}=\frac{AK\sin \beta}{BL\sin \alpha}$. Now, $AK=c \cdot \cos \alpha$ and $BL=c \cdot \cos \beta,$ so $\frac{AK\sin \beta}{BL\sin \alpha}=\frac{\sin \beta \cos \alpha}{\cos \beta \sin \alpha}=\frac{AN}{BN}$. So, $VN$ is bisector of $\angle AVB \implies \angle AVN=\angle BVN$. Since we now have two pairs of equal angles, that means that $\angle NVL=90^{\circ}$

This problem also appears in January, 2014 Crux Mathematicorum OC103.