TITLE: Geometry high school math competition question
QUESTION [5 upvotes]: Let $ABC$ be an equilateral triangle with side length $2$.
  Let the circle with diameter AB be $\gamma$.
  Consider the two tangents from $C$ to $\gamma$, and let the tangency point closer to $A$ be $D$.
  Find the area of triangle $CAD$.

I was able to figure out that $CD$ has to be $\sqrt2$. I can not figure out the height of triangle $CAD$.
I am trying to calculate the height from $D$ to $AC.$ Alternatively, if I could find the length of $AD$, then I should also be able to solve problem.

REPLY [3 votes]: \begin{align}
S_{ACD}&= 
S_{CDO}+S_{AOD}-S_{AOC}
,\\
S_{CDO}&=\tfrac12|CD|\cdot|OD|=\frac{\sqrt2}2
,\\
S_{AOD}&=\tfrac12|OA|\cdot|OD|\cdot\sin\angle AOD
=\tfrac12\sin\angle OCD
=\tfrac12\cdot\frac{|OD|}{|OC|}
=\frac{\sqrt3}6
,\\
S_{AOC}&=\tfrac12|OA|\cdot|OC|=\frac{\sqrt3}2
,\\
S_{ACD}&=\frac{\sqrt2}2+\frac{\sqrt3}6-
\frac{\sqrt3}2
=\frac{\sqrt2}2-\frac{\sqrt3}3
\approx .1297565117
.
\end{align}