Let $M$ be the midpoint of $AC$, also let $O'$ be the point on $AO$ such that $AH=AO'$ and $T"$ be the point such that $O'T"=OT$ and $AT"\perp T"H$. Hence $\triangle AMO\cong\triangle AT"H$, hence $AM=AT"$, which implies $\triangle AT"O'\cong\triangle AMH$, therefore $\angle O'AT"=\angle HAM$. But $\angle OAT=\angle HAM$, therefore $\angle OAT"=\angle OAT$. So $T"$ and $T$ are points such that $\angle OAT"=\angle OAT$ and $\angle AT"H=\angle ATH$, thus $T"=T$. Now $O'T=OT$ and $\angle TAO=\angle TAO'$, so $O'=O$. Therefore we conclude that $AO=AH$. Let $N$ be the midpoint of $AB$. We have $\triangle ANO\cong\triangle AT'H$, thus $OT'=HN=AN=\frac12 AB$, as desired.

Sorry, exist two points T'' which belong to the circle with diameter $AH$ and $O'T''=OT$ . I didn't understand your proof. Maybe enter in some details. Thank you. I have a proof with the power of a point w.r.t. circumcircle of $ABC$ .

From $\angle OAT"=\angle OAT$ we know that $A,T,T"$ are collinear. So $T$ and $T"$ are intersections of line $AT$ and circle with diameter $AH$, so they must coincide because the other intersection is point $A$.

Quote: Let $ABC$ be an acute triangle with the circumcircle $w=C(O,R)$ . Denote the midpoints $E$ , $F$ of $[AC]$ , $[AB]$ respectively, the projection $D$ of $A$ on $BC$ and the projections $X$ , $Y$ of $D$ on $AB$ , $AC$ respectively. Prove that $OX=AE\iff OY=AF\iff h_a=R$ . Proof of PP1. Since $\begin{array}{ccccc} DX^2=XA\cdot XB & ; & DX=h_a\cos B & ; & OE=R\cos B\\\ DY^2=YA\cdot YC & ; & DY=h_a\cos C & ; & OF=R\cos C\end{array}$ obtain that $\left\|\begin{array}{cccc} XA\cdot XB=R^2-OX^2 & \implies & OX^2=R^2-DX^2\\\ YA\cdot YC=R^2-OY^2 & \implies & OY^2=R^2-DY^2\end{array}\right\|$ . Thus, $OX=AE \iff R^2-AE^2=DX^2 $ $\iff OE=DX \iff h_a=R\ .$ $OY=AF \iff R^2-AF^2=DY^2 $ $\iff OF=DY \iff h_a=R\ .$ Remark. $h_a=R\iff bc=2R^2\iff \sin B\sin C=\frac 12$ .

Virgil wrote: $R^2-AE^2=DX^2 $ $\iff OE=DX$ How?

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