TITLE: How to calculate the area of $\triangle ABC$ when the distance from $BC$ to the circumcircle at $G$ is 10?
QUESTION [6 upvotes]: $\triangle ABC$ is right angle triangle and its circumcenter is $O$.
$G$ is a point where $BC$ is tangent to the incircle. The
perpendicular distance from $BC$ to circumcircle at $G$ is 10. How to
calculate the area of $\triangle ABC$?

I have tried to prove if the incenter, circumcenter and orthocenter are collinear but failed. I couldn't find what was special about the point $G$. What would be the correct approach to solve this problem?

REPLY [6 votes]: Let $|BC|=a$, $|AC|=b$, $|AB|=c$,
$|GE|=|DE|$.
The distances to the tangent point $G$ of the incircle are
\begin{align}
|BG|&=\tfrac12(a+c-b)
\tag{1}\label{1}
,\\
|CG|&=\tfrac12(a+b-c)
\tag{2}\label{2}
,
\end{align}
and by
the power of the point
$G$ w.r.t the circumcircle,
\begin{align}
|BG|\cdot|CG|&=|DG|\cdot|EG|=|DG|^2=100
\tag{3}\label{3}
,\\
|BG|\cdot|CG|&=\tfrac14(a+c-b)(a+b-c)
=\tfrac14(a^2-(c-b)^2)
=\tfrac14(b^2+c^2-(c-b)^2)
=\tfrac12\,bc
\tag{4}\label{4}
,
\end{align}
hence, the area of $\triangle ABC$ is $100$.