# How to calculate the area of $\triangle ABC$ when the distance from $BC$ to the circumcircle at $G$ is 10?

$$\triangle ABC$$ 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?

• Hint use the tangents . They are the key to the problem Aug 31, 2020 at 5:08
• Hint: Join $BD$ and $DC$ to get $BG\times GC=100$ Aug 31, 2020 at 5:15
• @Anand is the area 100? Aug 31, 2020 at 5:44

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 $$100$$.

• how did you get the lengths of BG and CG? Aug 31, 2020 at 7:51
• @C Squared: It's a very well-known property of the tangent points of the incircle in any triangle, which can be easily deduced. Consider all three tangent points. Let the distances from the vertices $A,B,C$ to any adjacent tangent point be $x,y$ and $z$. Then $x+y=c,y+z=a,z+x=b$, find $x,y,z$. Aug 31, 2020 at 8:09

Euler's theorem states that the distance d between the circumcentre and incentre of a triangle is given by $$d^{2}=R(R-2r)$$.

Let $$I$$ be a center of incircle. We have

$$OI^2 = IG^2 + OG^2$$ or $$OG^2 = OI^2 - IG^2 = R(R-2r)-r^2.$$

On other hand, we have $$OD^2 = DG^2 + OG^2$$ or $$R^2 = DG^2 + (R^2-2Rr - r^2)$$ Then $$DG^2 = r(2R+r) = 100.$$

Note that $$S_{ABC} = \frac{r(AB+BC+CA)}{2} = \frac{r(2r+4R)}{2}=r(2R+r).$$

So we have $$S_{ABC} = 100$$.