Find Length II

Let the center of the circle be the origin $O(0, 0)$ . Then the equation of the circle is $x^2+y^2 = 36$ . The $y$ -coordinate of $B$ is given by $3^2 + y_B^2 = 36$ $\implies y_B = - \sqrt{36-9} = -3\sqrt 3$ . The equation of chord $BC$ is given by:

$\begin{aligned} \frac {y+3\sqrt 3}{x-3} & = - \tan 60^\circ = - \sqrt 3 \\ y + 3\sqrt 3 & = 3\sqrt 3 - \sqrt 3 x \\ \implies y & = - \sqrt 3 x \end{aligned}$

Since the $y$ -intercept of chord $BC$ is 0, this means that it passes through the origin $O(0,0)$ or the center of the circle. Therefore $BC$ , which is a side of equilateral $\triangle ABC$ , is a diameter of the circle and has a length of $\boxed {12}$ .

Let $DE$ cross $BC$ at $X$ .

Triangle $BXF$ is a $(60*, 30*, 90*)$ triangle, so $|BX| = 6$ . Hence $BX$ is a radius, $BC$ is a diameter, and the side length is $12$ .

Mark intersection of CB and ED as O.

$OB = \frac{FB}{ cos(60)}= \frac{3}{\frac{1}{2}} = 6$

Since OB = R = 6, than O must be the center of the circle. $BC = 2 R$

$AB = BC = AC$

$BC = 2*6 = \boxed{12}$

Let $G$ be the intersection of $\overline{AB}$ with the circumference and $H$ to the intersection between $\overline{ED}$ and $\overline{BC}$ . As $\overline{GF} \cong \overline{FB}$ , $\bigtriangleup FBH$ is equilateral of side lenght $6$ , the same as the radius, then $\overline{BC}$ is diameter and its lenght is $12$