Geometry high school math competition question 
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.
 A: 
\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} 
A: Let $O$ be a center of the circle $\Gamma$.
Thus, by the Pythagoras's theorem 
$$CD^2=CO^2-DO^2=\left(\sqrt3\right)^2-1^2=2$$ and $$CD=\sqrt2.$$
Id est, $$S_{\Delta ACD}=\frac{1}{2}CD\cdot AC\sin\measuredangle ACD=$$
$$=\frac{1}{2}\cdot\sqrt2\cdot2\sin\left(\arctan\frac{1}{\sqrt2}-30^{\circ}\right)=$$
$$=\sqrt2\left(\sin\arctan\frac{1}{\sqrt2}\cdot\frac{\sqrt3}{2}-\cos\arctan\frac{1}{\sqrt2}\cdot\frac{1}{2}\right)=$$
$$=\sqrt2\left(\frac{\frac{1}{\sqrt2}}{\sqrt{1+\left(\frac{1}{\sqrt2}\right)^2}}\cdot\frac{\sqrt3}{2}-\frac{1}{\sqrt{1+\left(\frac{1}{\sqrt2}\right)^2}}\cdot\frac{1}{2}\right)=\frac{1}{\sqrt2}-\frac{1}{\sqrt3}.$$
There is a solution without trigonometry.
Let $DK$ be an altitude of $\Delta ADO$.
Thus, since $\Delta OKD\sim\Delta CDO,$ we obtain:
$$\frac{DK}{DO}=\frac{DO}{CO}$$ or
$$\frac{DK}{1}=\frac{1}{\sqrt3},$$ which gives $$DK=\frac{1}{\sqrt3}$$ and
$$S_{\Delta ACD}=S_{\Delta ADO}+S_{\Delta OCD}-S_{\Delta ACO}=$$
$$=\frac{1}{2}\cdot\frac{1}{\sqrt3}\cdot1+\frac{1}{2}\cdot\sqrt2\cdot1-\frac{1}{2}\cdot\sqrt3\cdot1=\frac{1}{\sqrt2}-\frac{1}{\sqrt3}.$$
