Selection and Training Session

Let $l_a$ be the bisectrix of $\angle A$. Then $l_a\perp AT$ and $PH\parallel l_a$ (as, by condition, $PH\perp AT$). Further, $$\angle (BH,HP)=\angle (BH,AC)+\angle (AC,HP)=90^{\circ }+\angle (AC,l_a)=90^{\circ }+\angle (l_a,AB)=90^{\circ }+\angle (HP,AB)=\angle (HP,AB)+\angle (AB,CH)=\angle (HP,CH).$$ Hence, $HP$ is the bisectrix of $\angle BHC$.

Let $AH$ meet the circumcircle of $△ABC$ at $D_1\ne A$. We have $\angle D_{1}BC=90^{\circ }-\angle C=\angle HBC$ and, similarly, $\angle D_{1}CB=\angle HCB$. Then the triangles $BHC$ and $B{D}_{1}C$ are equal, Hence, $D_{1}P$ is the bisectrix of $\angle B{D}_{1}C$. Therefore, $D_{1}P$ intersects the circumcircle at the middle of the arc $BAC$, i.e. at the point $T$. So, $D_1=D$ and $AD\perp BC$. Thus the statement follows from the well-known

Lemma. The diagonals of the quadrilateral $ABCD$ are perpendicular if and only if $A{B}^2+C{D}^2=A{C}^2+B{D}^2$.