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

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

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 $\angle AHO = 90^\circ$.