Find the length!

We join $B,T$ and $T,C$ . Thus in quadrilateral $TBHC$ we have the diagonals bisecting each other whcih implies $TAHC$ is a parallelogram. Thus, $AH=TC$ and $TC \parallel AC$ which means $\angle TCA = 90^{\circ}$ . Now, let $O$ be the circumcentre of $\triangle ABC$ and as $\triangle ABC$ is acute, $O$ lies inside the triangle. WE drop $OX \perp AC$ .

Now, $OX = \frac{1}{2} \times AH = \frac{1}{2} \times TC$ . And along with that $TC \parallel OX$ and $X$ is the midpoint of $AC$ . This implies $O$ is the midpoint of $AT$ and thus $AT$ is the diameter of the circumcircle of $\triangle ABC$ . Finally we use the extended Law of Sines in $\triangle ABC$ which yields

$\dfrac{BC}{\sin \angle A} = 2R$ , where $R$ is the circumradius of $\triangle ABC$

$\implies$ $\dfrac{5}{\sin 30^{\circ}} = AT$

$\implies$ $AT = \boxed{10}$

This question uses a fact that orthocentre reflected about a midpoint always lies on a circumcircle, on a diameter of circumcircle with opposite vertex as the other end of the diameter.

Let $R$ denote radius of circumcircle, $O$ its circumcentre. In this case it means that $AT=2R$ . By inscribed angle theorem $\angle BOC = 2 \times \angle BAC = 60 \Rightarrow \triangle BOC$ is equilateral $\Rightarrow R=BC=5 \Rightarrow AT=\boxed{10}$ .

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