Why is this a right angle in the mach cone? 
This and similar depictions are used when deriving the equation for the Mach cone angle $\theta$.
It is usually stated that using algebra we can conclude that
$$\sin\theta=\frac{ct}{vt},$$
and thus
$$θ=\arcsin \left(\frac{c}{v}\right).$$
However, I noticed that that assumes a right angle between $ct$ and the wavefront, which is never mathematically proven.
So, can someone explain why there is a right angle between $ct$ and the wavefront?
 A: This is a standard theorem in Euclidean geometry: the tangent to a circle is perpendicular to the radius at the point of tangency. The proof is a simple proof by contradiction.
Given: a circle $C$, a line tangent to the circle $T$, and a radius, $R$, from the center of the circle to the point of tangency.
Assume: $R$ is NOT perpendicular to $T$
Then there is some other line segment, $S$, from the center of $C$ which is perpendicular to $T$. Since $S$ is perpendicular to $T$ that makes $R$ the hypotenuse of a right triangle. Therefore the length of $R$ is greater than the length of $S$, so $|R|>|S|$.
Now, except for the point of tangency all other points on $T$ are outside of $C$. Since $C$ is defined as all points a distance of $|R|$ away from the center, all points outside of $C$ are a distance $x>|R|$ away from the center. Since $S$ connects the center to such a point we have $|S|>|R|$.
But $|S|>|R|$ contradicts $|R|>|S|$ so our assumption is false. Therefore $R$ is perpendicular to $T$. QED
