Proving $|CH| = 2|OM|$ of $\triangle ABC$ of which $O$ is the circumcenter and $H$ is the orthocenter. Let $ABC$ be a triangle as shown in the figure below, where $O$ is its circumcenter and $H$ is its orthocenter. $AB$ is the opposite side of the climax point $C$, and $OM$ is perpendicular to $AB$. 
How to prove that $|CH| = 2|OM|$?

$G$ is the centroid of the triangle. As the circumcenter, centroid and orthocenter (respectively $O$, $G$, and $H$) of the circle is collinear, we can get two specific triangle $CGH$ and $OMG$. 
Showing both the triangle similar to each other by their respective angle, we can get the proportion of their 2 different length. If we consider the point $G$, then we can get $|CG|:|GM| = 2:1$. Thus we can easily prove this. 
However, we can reverse the proof showing that $|CH|$ is equal to $2|OM|$ (It is included in our textbook as a corollary or something other). If we go with this process, it will lead to something contradictory. 
Because of claiming the $G$ as centroid of the triangle, we have already used the above condition which needs to be proved. 
So, is there any other method to prove that $|CH|=2|OM|$?
 A: 
The most ellegant proof (at least in my opinion) is vector based:
Vectors $\vec{CH}$ and $\vec{OM}$ are parallel as well as vectors $\vec{AH}$ and $\vec{ON}$ and therefore:
$$\vec{OM}=\alpha\vec{CH}$$
$$\vec{ON}=\beta\vec{AH}$$
...for some real values of $\alpha,\beta$. 
Segment MN connects midpoints of sides $AB$ and $BC$ and therefore:
$$\vec{MN}=\frac12\vec{AC}=\frac12(\vec{AH}+\vec{HC})=\frac12\vec{AH}-\frac12\vec{CH}\tag{1}$$
On the other side it is obvious that:
$$\vec{MN}=\vec{MO}+\vec{ON}=-\vec{OM}+\vec{ON}=-\alpha\vec{CH}+\beta\vec{AH}\tag{2}$$
From (1) and (2) you get that:
$$\frac12\vec{AH}-\frac12\vec{CH}=-\alpha\vec{CH}+\beta\vec{AH}$$
$$(\frac12-\beta)\vec{AH}+(\alpha-\frac12)\vec{CH}=0\tag{3}$$
Vectors $\vec{AH}$ and $\vec{CH}$ are not parallel! If you multiply these vectors with some real factors different from zero, you still get vectors that are not parallel (vectors are just stretched). Sum of two vectors that are not parallel cannot be zero. So (3) can be valid only if both scalars are equal to zero:
$$\frac12 - \beta =0,  \ \ \alpha-\frac12=0$$
...which simply means that $\alpha=\beta=\frac12$. Consequentially:
$$\vec{OM}=\frac12\vec{CH}$$
$$OM=\frac12 CH$$
A: Do you know Hamilton theorem? It says  $$\vec{OH}= \vec{OA}+\vec{OB}+\vec{OC} $$
So $$\vec{CH}=\vec{OH}- \vec{OC}= \vec{OA}+\vec{OB}   =2\vec{OM}   $$
