Relation between cotangent and length of perpendicular 
I am trying to prove this relation for any acute triangle:
$$\frac{\cot(C)}{2}=\frac{OR}{AB}$$
where

*

*$O$ is the center of the circumscribed circle,

*$C$ is the angle opposite the side $AB,$

*$R$ is the foot of a perpendicular dropped from $O$ to the side $AB.$
This relation is used in this paper (page 6) equations 6 and 7.
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I think the left side of the equation is actually:
$\frac{\cot(C)+\cot(X)}{2}$ but since $X=\pi/2, \cot(X)=0$
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I tried to use the law of cotangents https://en.wikipedia.org/wiki/Law_of_cotangents but was not successful because of the half-angle.
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 A: There's this theorem
It states that in any circle the angle made by an arc at center is twice to the angle made by the same arc at any point on the circle (other than on the arc itself).
So, for the circumcircle of the triangle $ABC$, point $O$ would be the center.
Imagine an arc made up of points $A$ and $B$. It makes $\angle C$ at a point on the circle (since point $C$ lies on the circumcircle)
It also makes an angle $\angle AOD$ at $O$, which is the center of the circumcircle.
So, from the above mentioned theorem:
$C =  \frac{\angle AOD}{2}$
from figure we can see that
$\angle AOD = 2\angle AOR$
From above two statements, we have
$C = \angle AOR$
Since $OR$ is perpendicular to $AR$ we have $\cot(\angle AOR)$ = $OR \over AR$
Since, $C = \angle AOR$,
$\cot C$ = $OR \over AR$
And, $AR$ is twice of $AB$
So, $\cot C$ = $2OR \over AB$
or
$${\cot C \over 2} = {OR \over AB}$$
Hence, proved.
This is true as long as angle C is less than right angle (from above theorem). Circumcenter will be outside for obtuse triangles.
