Prove that the intersection of a sphere and plane is a circle 
Prove that if a plane has two distinct common points with a sphere centered at $O$, then the intersection is a circle with some center $O_{1}$, where segment $OO_{1}$ is orthogonal to the plane.

I chose two points $A$ and $B$ that are on the intersection of the plane and the sphere. They are the same distance apart since $A$, $B$ are in the sphere. The two triangles share a leg $OO_{1}$. Since the problem says $OO_{1}$ is orthogonal that means theirs ninety degrees on each side, which means the angles are congruent. 
However, I can not figure out how to show $AO_{1}$ and $BO_{1}$ are congruent. I also can't figure out how to show that no matter what points you pick in the intersection they will be the same distance away from $O_{1}$. 
This is a projective geometry problem. 
 A: You have side, side, angle. This determines (since your angle is right) the other remaining side of the triangle through the Law of Cosines. Thus, $AO_1$ and $BO_1$ are of equal length.
In particular, your triangles are both right, with a hypotenuse of the same length, and a shared leg. Pythagorean Theorem (a special case of the Law of Cosines) tells us that the other legs of the two triangles have the same length, as well.
A: How about this approach (I'm assuming that we are working in $\mathbb{R}^3$)?
Let the plane be $H = \{x | \langle x , u_1 \rangle = c \}$, where $\|u_1\| = 1$. Let the sphere be $C = \{ x | \|x\| = r \}$. Augment $u_1$ with $u_2,u_3$ such that they form an orthonormal basis. Note that $x \in C$ iff $\sum_k \langle x , u_k \rangle^2 = r^2$. Consequently, $x \in H \cap C$ iff $\langle x , u_1 \rangle = c$ and $\langle x , u_2 \rangle^2 + \langle x , u_3 \rangle^2 = r^2 -c^2$. It follows that $H \cap C = \{ c u_1 + \alpha_2 u_2 + \alpha_3 u_3 | \alpha_2^2+\alpha_3^2 = r^2-c^2\}$.
Since $H \cap C$ is non-empty, we have $r^2 \ge c^2$, and since $|H \cap C|\ge 2$, we have $r^2 > c^2$ (otherwise there would just be a single point).
Hence $H \cap C$ is a circle of radius $r^2-c^2$ centered at $c u_1$, and the line joining the center of the sphere to the center of the circle (ie, $\operatorname{sp} \{ u_1 \}$) is perpendicular to $H$ (or more correctly, perpendicular to $H-\{c u_1\}$).
