APMO

Suppose, without loss of generality, that $B$ and $C$ lie on the same side of line $OH$. Such line is the Euler line of $ABC$, so the centroid $G$ lies on this line.



Let $M$ be the midpoint of $BC$. Then the distance between $M$ and the line $OH$ is the average of the distances from $B$ and $C$ to $OH$, and the sum of the areas of triangles $BOH$ and $COH$ is $$[BOH]+[COH]=\frac{OH\cdot d(B,OH)}{2}+\frac{OH\cdot d(C,OH)}{2}=\frac{OH\cdot 2d(M,OH)}{2}.$$ Since $AG=2GM$, $d(A,OH)=2d(M,OH)$. Hence $$[BOH]+[COH]=\frac{OH\cdot d(A,OH)}{2}=[AOH],$$ and the result follows.

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Alternative solution.

One can use barycentric coordinates: it is well known that $$\begin{array}{c}A=(1:0:0),B=(0:1:0),C=(0:0:1),\\ O=(\sin 2A:\sin 2B:\sin 2C)\text{and}H=(\tan A:\tan B:\tan C).\end{array}$$ Then the (signed) area of $AOH$ is proportional to $$\left|\begin{array}{ccc}1& 0& 0\\ \sin 2A& \sin 2B& \sin 2C\\ \tan A& \tan B& \tan C\end{array}\right|$$ Adding all three expressions we find that the sum of the signed areas is a constant times $$\left|\begin{array}{ccc}1& 0& 0\\ \sin 2A& \sin 2B& \sin 2C\\ \tan A& \tan B& \tan C\end{array}\right|+\left|\begin{array}{ccc}0& 1& 0\\ \sin 2A& \sin 2B& \sin 2C\\ \tan A& \tan B& \tan C\end{array}\right|+\left|\begin{array}{ccc}0& 0& 1\\ \sin 2A& \sin 2B& \sin 2C\\ \tan A& \tan B& \tan C\end{array}\right|.$$ By multilinearity of the determinant, this sum equals $$\left|\begin{array}{ccc}1& 1& 1\\ \sin 2A& \sin 2B& \sin 2C\\ \tan A& \tan B& \tan C\end{array}\right|,$$ which contains, in its rows, the coordinates of the centroid, the circumcenter, and the orthocenter. Since these three points lie on the Euler line of $ABC$, the signed sum of the areas is $0$, which means that one of the areas of $AOH$, $BOH$, $COH$ is the sum of the other two areas.