I have this picture where the textbook of my students of 17 years old say to find the area of the quadrilater $AOBC$ (with $|AO|=|OB|=r$ the radius of the circle with the center in $O$), your perimeter and the area of mixtilinear triangle $\triangle ACB$ knowing that the angle $\angle ACB=\pi/3$. In particular the textbook say that to take two point on the circle $A$ and $B$ and to draw the straight tangent lines that they meet in a point $C$ forming an angle of $\angle ACB=\pi/3$.
The problem was solved in class but I made some considerations. If I draw everything with Geogebra or any other software I observe that the triangle $\triangle ABC$ is equilater: in fact $CH$ cut $AB$ in two equal parts and the median $CH$ is also the bisector of the angle $\angle ACB$. Hence $\angle CBH=\angle BAC=\pi/3$.
If I would have drawn the triangle wrongly, with an angle of $\pi/3$ and didn't notice that it is an equilateral triangle, is there a quick way to prove it?
Trying to be rigorous if I have two tangent lines to the circumference, at two points $AB$ and $B$ that, intersected, form an angle of $\pi/3$, how do I guess that triangle $ABC$ is equilateral? If the angle had been any other amplitude it might as well have been isosceles or scalene.
I have thought to use a circle to center $O(0,0)$ and radius $r$
$$x^2+y^2=r^2$$ and $$\tan \angle ACB=\frac{\pi}3=\left|\frac{m-m'}{1+mm'}\right|$$ where $m$ it is the slope of the tangent in the point $A$ and $m'$ in the point $B$, but I not know the coordinates of $A$ and $B$.
Triangles $$ and $$ are equal by hypotenuse-leg congruence, hence $=$. But base angles of an isosceles triangle are equal, hence: $$ \angle CAB=\angle CBA ={180°-60°\over2}=60°. $$