Prove that the sum of areas of triangles $AOH$ and $BOH$ equals the area of triangle $COH$. Let $O$ be the circumcenter and $H$ the orthocenter of an
acute triangle $ABC$. Prove that the area of one of the triangles $AOH, BOH,$ and $COH$ is
equal to the sum of the areas of the other two.

In this figure, we want to prove $[AOH]+ [BOH]= [COH]$. Where $[.] $ is the area of a given shape.
Let $A',B'$ and $C'$ be the othogonal projections of the points $A,B,C$, respectively onto the line $OH$. Now note that
$$[AOH]+ [BOH]= [COH]\iff $$
$$AA'\cdot OH+BB'\cdot OH=CC'\cdot OH$$
$$\iff AA'+BB'=CC'$$
Any idea how to show this? By the way, I know this problem already exists here but I don't want to solve it using analytical techniques (Vectors, coordinates).
 A: Let $(d)$ be any line through the centroid $G$.
Let $A'$ be the feet of the perpendicular from $A$ to $(d)$, $(AC)\cap (d)=A'',$ $
\ (BC)\cap(d)=B''$, and $M$ be the midpoint of side BC.
Define $B', C'$ similarly.
Wlog, $A$ and $B$ lie on the same side of $(d)$
Note, by Thales's Theorem:
$$\frac{BB'}{CC'}=\frac{BB''}{B''C}$$ $$\frac{AA'}{CC'}=\frac{AA''}{A''C}$$
Thus, what we want is equivalent to:
$$\frac{AA''}{CA''}+\frac{BB''}{CB''}=1$$
which we obtain directly by applying Menelaus's theorem, in triangle $\triangle ACM$ and collinearity of $G, A''$ and $B''$.
A: Here is a possibility to proceed geometrically.
We isolate the Euler line as the main ingredient in the proof, then use the fact that the centroid is a point on it.
Picture of the situation:

Here, $G,H,O$ are respectively the centroid, the orthocenter, and the circumcenter of $\Delta ABC$. Thle points $G,H,O$ are on a line, the Euler line of the triangle.
For each vertex $X$ among $A,B,C$ i have denoted

*

*by $X_G$ the intersection of $XG$ with the opposite side,

*by $X_H$ the intersection of $XH$ with the opposite side,

*by $X_E$ the intersection of the Euler line with the opposite side,

*by $X'$ the projection of $X$ on the Euler line, as in the question.

In the picture the vertex $A$ is chosen so that $A_E$ is not on the opposite side (seen as a segment).
We apply the theorem of Menelaus in $\Delta ABA_G$, $\Delta ACA_G$ w.r.t. the Euler line:
$$
\begin{aligned}
1 &= 
\frac{C_EB}{C_EA}\cdot
\frac{GA}{GA_G}\cdot
\frac{A_EA_G}{A_EB}
=
\frac{B'B}{A'A}\cdot 
2\cdot
\frac{A_EA_G}{A_EB}
\ ,
\\
1 &= 
\frac{B_EC}{B_EA}\cdot
\frac{GA}{GA_G}\cdot
\frac{A_EA_G}{A_EC}
=
\frac{C'C}{A'A}\cdot 
2\cdot
\frac{A_EA_G}{A_EC}
\ ,
\end{aligned}
$$
and from here:
$$
\frac{B'B}{A'A}
+
\frac{C'C}{A'A}
=
\frac 12\left(
\frac{A_EB}{A_EA_G}+
\frac{A_EC}{A_EA_G}
\right)
=1
\ ,
$$
since $A_G$ is the mid point of $BC$.
