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.
On
\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}
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 $$S_{\Delta ACD}=S_{\Delta ADO}+S_{\Delta OCD}-S_{\Delta ACO}=$$ $$=\frac{1}{2}\cdot\frac{1}{\sqrt3}\cdot1+\frac{1}{2}\cdot\sqrt2\cdot1-\frac{1}{2}\cdot\sqrt3\cdot1=\frac{1}{\sqrt2}-\frac{1}{\sqrt3}.$$