For an isosceles prism of angle $A$ and refractive index $\mu$, it is found that the angle of minimum deviation ${\delta }_m=A$. Which of the following options is/are correct?

The minimum deviation produced by a prism ${\delta }_m=2i-A=A$

$$\begin{gathered} \Rightarrow i_1=i_2=A \text{and} r_1=r_2=\frac{A}{2} \\ \Rightarrow r_1=\frac{i_1}{2} \end{gathered}$$

Now, using Snell's law,

$\begin{array}{rcl}\sin A& =& \mu \sin \left(\frac{A}{2}\right)\\ & \Rightarrow & \mu =2 \cos \left(\frac{A}{2}\right)\end{array}$

For this prism when the emergent ray at the second surface is tangential to the surface

$$\begin{gathered} i_2=\frac{\mathrm{\pi }}{2} \\ {r}_2={\theta }_c \\ {r}_1=A-{\theta }_c \end{gathered}$$

$$\begin{gathered} \sin i_1=\mu \sin (A-{\theta }_c) \\ \Rightarrow i_1={\sin }^{-1}\left[\sin A\sqrt{4{\cos }^2\left(\frac{A}{2}\right)-1}-\cos A\right] \end{gathered}$$

For minimum deviation through isosceles prism, the ray inside the prism is parallel to the base of the prism if $\angle B=\angle C$.

But it is not necessarily parallel to the base if,

$\begin{array}{rcl}\angle A& =& \angle B\\ & \Rightarrow & \angle A=\angle C\end{array}$