Let $\Sigma$ be a sphere in 3 dimensional Euclidean space, and let $\Pi_1$ and $\Pi_2$ be planes in this space that intersect $\Sigma$. Define $C_i := \Sigma\cap\Pi_i$, $i \in \{1,2\}$. Suppose $C_1\cap C_2$ consists precisely of two distinct points, say $a$ and $b$. How can it be shown that the angle between the tangents to $C_1$ and $C_2$ at $a$ is the same as the angle between the tangents to $C_1$ and $C_2$ at $b$?
* This is a rewording of problem 5.11 on p. 93 pf A. I. Markushevich's Theory of Functions of a Complex Variable, 2nd edition (three volumes in one), Chelsea Publishing Company, 1977.
My attempt at a solution
It is given that $a, b \in \Pi_1 \cap \Pi_2$ and that $a\neq b$. Hence the unique line $L$ that passes through $a$ and $b$ is contained in $\Pi_1\cap\Pi_2$. It is further given that $\Pi_1\neq\Pi_2$. Hence $\Pi_1\cap\Pi_2 = L$.
Now, here's the part I'm not sure about: intuitively, it feels like the tangent line $T^a_1$ to $C_1$ at $a$ is orthogonal to $L$, and likewise the tangent line $T^a_2$ to $C_2$ at $a$; and that the tangent lines $T^b_1$ and $T^b_2$ to $C_1$ and $C_2$, respectively, at $b$ are likewise orthogonal to $L$.
Assuming this to be the case, the angle $\alpha_1$ between $T^a_1$ and $T^a_2$ is the angle between $\Pi_1$ and $\Pi_2$, and the same is true of the angle $\alpha_2$ between $T^b_1$ and $T^b_2$. Hence $\alpha_1 = \alpha_2$, as was to show.