Estonian Math Competitions

Let $O^{\prime }$ be the other endpoint of the diameter of the circumcircle of the triangle $ABC$ drawn from the vertex $A$ and let $H^{\prime }$ be the reflection of the point $H$ through the line $BC$ (Fig. 35). Then $HD=H^{\prime }D$, whence the condition $\frac{AH}{HD}=2$ is equivalent to the condition $\frac{AH}{A{H}^{\prime }}=\frac{1}{2}=\frac{AO}{A{O}^{\prime }}$. Thus $\frac{AH}{HD}=2$ if and only if $OH\parallel O^{\prime }{H}^{\prime }$.





It is known that $H^{\prime }$ lies on the circumcircle of the triangle $ABC$. Thus by Thales' theorem, $\angle A{H}^{\prime }{O}^{\prime }=90^{\circ }$. Hence the condition $OH\parallel O^{\prime }{H}^{\prime }$ is equivalent to the condition $\angle AHO=90^{\circ }$. Consequently, $\frac{AH}{HD}=2$ if and only if $\angle AHO=90^{\circ }$.

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Alternative solution.

W.l.o.g., assume $AC\le AB$. Let $G$ be the centroid of the triangle $ABC$ and $K$ be the midpoint of the side $BC$ (Fig. 36). Then $\frac{AG}{GK}=2$. Thus $\frac{AH}{HD}=2$ if and only if $\frac{AH}{HD}=\frac{AG}{GK}$, which in turn is valid if and only if $HG\parallel DK$. The latter condition is equivalent to $\angle AHG=\angle ADK=90^{\circ }$. But $H,G$ and $O$ lie on the same line (Euler line). Thus $\angle AHG=90^{\circ }$ if and only if $\angle AHO=90^{\circ }$. Hence $\frac{AH}{HD}=2$ if and only if $\angle AHO=90^{\circ }$.