The locus of the middle points of the chords of the circle $$C_1: (x-4)^2 + (y-5)^2 = 4$$ which subtend an angle $$\theta_i$$ at the centre of the circle $$C_i$$, is a circle of radius $$r_i$$. If $$\theta_1 = \frac{\pi}{3}$$, $$\theta_3 = \frac{2\pi}{3}$$ and $$r_1^2 = r_2^2 + r_3^2$$, then $$\theta_2$$ is equal to

$$r_i = R \cos\left(\frac{\theta_i}{2}\right)$$

Given the circle equation $$C_1 : (x - 4)^2 + (y - 5)^2 = 4$$, its radius is $$R = \sqrt{4} = 2$$.

$$\Rightarrow r_i = 2 \cos\left(\frac{\theta_i}{2}\right)$$

$$r_1^2 = r_2^2 + r_3^2$$

$$\Rightarrow \left[2 \cos\left(\frac{\theta_1}{2}\right)\right]^2 = \left[2 \cos\left(\frac{\theta_2}{2}\right)\right]^2 + \left[2 \cos\left(\frac{\theta_3}{2}\right)\right]^2$$

$$\Rightarrow 4 \cos^2\left(\frac{\theta_1}{2}\right) = 4 \cos^2\left(\frac{\theta_2}{2}\right) + 4 \cos^2\left(\frac{\theta_3}{2}\right)$$

$$\Rightarrow \cos^2\left(\frac{\theta_1}{2}\right) = \cos^2\left(\frac{\theta_2}{2}\right) + \cos^2\left(\frac{\theta_3}{2}\right)$$

Given $$\theta_1 = \frac{\pi}{3}$$ and $$\theta_3 = \frac{2\pi}{3}$$:

$$\Rightarrow \frac{3}{4} = \cos^2\left(\frac{\theta_2}{2}\right) + \frac{1}{4}$$

$$\Rightarrow \cos^2\left(\frac{\theta_2}{2}\right) = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$$

$$\Rightarrow \cos\left(\frac{\theta_2}{2}\right) = \frac{1}{\sqrt{2}}$$

$$\Rightarrow \frac{\theta_2}{2} = \frac{\pi}{4}$$

$$\Rightarrow \theta_2 = \frac{\pi}{2}$$