Homework hint: use dot product to show inscribed angle of a semicircle is a right angle I'm currently stuck on a homework problem from MIT OCW 18.02 multivariate calculus.
The problem is in a section on dot products, and asks to use vector methods (without components) to show that the inscribed angle of a semicircle is a right angle.
I have been toying around with it the past couple of days. The first thing I did was sketch it out, and I quickly came to a solution by summing up the angles in my sketch. (I drew two triangles and determined the relationship between the angles, which led me to deduce the inscribed angle was $\frac{\pi}{2}$)
But that solution doesn't feel in keeping with the rest of the problems, which makes me wonder what I missed.
I know I could just peek the answer, but I also know that wouldn't really help. I would prefer a gentle nudge in the right direction instead. Can you provide a hint that may point me in the direction I need to think?
 A: Hint :
Since vectors encode same information as Geometry, we make use of same triangles in (your) Euclidean geometry proof.
Fix the center of semicircle as the origin. Let $A,B$ be the position vectors of end points of the base diameter and let $P$ be the position vector of any point on the semicircle. We have to prove
$$(P-A)\cdot (P-B)=0$$
Suppose $A=\vec{a}$, what would be $B$? Let $P=\vec{r}$.
Now that we know two sides of each triangle, we can find out the third.
Also note that $|OP|=|OA|=|OB|$. Now we can find out $PA \cdot PB$ and see if it evaluates to zero.
A: One can compute the dot product with a rotated coordinate system:

With the tangent and diameter to the circle at the inscribed angle as axes, we get
$$
\begin{align}
(B-C)\cdot(A-C)
&=\overbrace{\overbrace{\ (r+x)\ }^{B-C}\overbrace{\ (r-x)\ }^{A-C}}^\text{diametric}+\overbrace{\overbrace{\quad y\quad\phantom{()}}^{B-C}\overbrace{\quad(-y)\quad}^{A-C}}^\text{tangential}\\[6pt]
%&=r^2-x^2-y^2
\end{align}
$$
However, I am not sure if this avoids the "without components" restriction.

Evidently, the answer above is not what they wanted for "without components".  Here is a diagram of the answer given

