Let $\overline{CH} \cap \overline{AB} = F$ and $\overline{BH} \cap \overline{AC} = E$. Radical axis theorem on $\odot BDHE$, $\odot BDT$ and $\odot AHEF \Rightarrow \overline{CT}$ is tangent to both $\odot BDT$ and $\odot AHEF$. $\overline{CT} $ tangent to $\odot AHEF \Rightarrow$ the polar of $C $ with respect to $\odot AHEF $ passes through $T $. Brokard on $AHEF \Rightarrow$ the polar of $C $ with respect to $\odot AHEF $ passes through $B $. Since $\overline{BT} $ is the polar of $C $ with respect to $\odot AHEF $ and $N $ is the center of $\odot AHEF $ we conclude that $\overline{CN} \perp \overline{BT}$.

Let $BH\cap AC=K$suppose $l$ is the common tangent of $\odot(BTC)$ and $\odot(N,AN)$ Note that $A,B,D,K$ are cyclic By root axis theorem,$AK,BD,l$ are concurrent Which means that $l$ go through $C$ So $\angle NTC=90^{\circ}=\angle ADC$ Hence $\angle TBD=\angle CTD=\angle CND$,$\angle TBD+\angle NCB=90^{\circ}\Rightarrow BT\bot CN$

Clearly $C$ lies on the radical axis of $(AH)$ and $(BDT)$. We now wish to show there exists a real number $r$ such that $C$ and $N$ are on the radical axis of the circumference of radius $0$, $T$, and the circumference centered at $B$ with radius $r$. If we manage to show this, by the fact that the radical axis is perpendicular to the line of the center of the circumferences, then we would be done. So we wish to show that there exists a real number $r$ such that $CT^2=CB^2-r^2$ and $NT^2=NB^2-r^2$. This is equivalent to showing that $BN^2-NT^2=CB^2-CT^2$. Using that $CT^2=CB\cdot CD$, then the $RHS$ is equal to $BD\cdot BC=Pow_B(AH)$, which is precisely the $LHS$, implying that we are done.

aqwxderf wrote: Let $\overline{CH} \cap \overline{AB} = F$ and $\overline{BH} \cap \overline{AC} = E$. Radical axis theorem on $\odot BDHE$, $\odot BDT$ and $\odot AHEF \Rightarrow \overline{CT}$ is tangent to both $\odot BDT$ and $\odot AHEF$. $\overline{CT} $ tangent to $\odot AHEF \Rightarrow$ the polar of $C $ with respect to $\odot AHEF $ passes through $T $. Brokard on $AHEF \Rightarrow$ the polar of $C $ with respect to $\odot AHEF $ passes through $B $. Since $\overline{BT} $ is the polar of $C $ with respect to $\odot AHEF $ and $N $ is the center of $\odot AHEF $ we conclude that $\overline{CN} \perp \overline{BT}$. Same Solution (not surprised though) To Prove: $CN \perp BT$ $$\text{Pow}_{\odot(BDT)}(C)=CE \cdot CA=CB \cdot CD =\textit{Pow}_{\odot(BDEA)}(C) $$so $C$ lies on radax which is common tangent at $T$. This gives polar of $C$ w.r.t $\odot(AH)$ to be $T$. By Brokard's theorem on $\odot(AH)$, the polar of $C$ w.r.t $\odot (AH)$ is $B$. Because $BT$ is polar of $C$ w.r.t $\odot(AH)$, we are done as $N$ is the center.

Let the feet of the altitudes from $B$ and $C$ onto $AC$ and $AB$ be $E$ and $F$ respectively. Further let $t$ be the common tangent of $(BDT)$ and $(AEF)$ at $T$. From the radical centre theorem on circles $(AEF)$, $(BDT)$, $(ABDE)$ we have that $AE$, $BD$ and $t$ concur. Therefore, $C \in t$ and therefore $T$ is a point such that $CT$ is tangent to $(AEF)$. Now it suffices to prove that $BT$ is the polar of $C$ with respect to circle $(AEF)$. But $T$ lies on this polar of $C$ by definition, so it suffices to show that $B$ also lies on this polar. But this is true as by Brokard's theorem, $\Delta BCX$ is self-polar, where $X = AH \cap EF$. $\square$ Edit: FIRST EVER PROJECTIVE SOL LETS GOOOOOOO

AshAuktober wrote: Let the feet of the altitudes from $B$ and $C$ onto $AC$ and $AB$ be $E$ and $F$ respectively. Further let $t$ be the common tangent of $(BDT)$ and $(AEF)$ at $T$. From the radical centre theorem on circles $(AEF)$, $(BDT)$, $(ABDE)$ we have that $AE$, $BD$ and $t$ concur. Therefore, $C \in t$ and therefore $T$ is a point such that $CT$ is tangent to $(AEF)$. Now it suffices to prove that $BT$ is the polar of $C$ with respect to circle $(AEF)$. But $T$ lies on this polar of $C$ by definition, so it suffices to show that $B$ also lies on this polar. But this is true as by Brokard's theorem, $\Delta BCX$ is self-polar, where $X = AH \cap EF$. $\square$ Edit: FIRST EVER PROJECTIVE SOL LETS GOOOOOOO Bro we had the same solution at literally the same time, so nice

L13832 wrote: Bro we had the same solution at literally the same time, so nice ikr so cool!!