Prove radius chord theorem without using congruent traingles

Question

Suppose that $P(a,b)$ and $Q(c,d)$ are two points on the unit circle $x^2 + y^2 = 1$, and let $M$ be the midpoint of chord $PQ$. (Without using congruent triangles), prove that $OM$ is perpendicular to $PQ$. (Here $O$ is the origin, which is the centre of the circle).

Im not sure where to begin with this problem. I was thinking of trying to prove that the product of the slopes is $-1$, though as $P$ and $Q$ may have any coordinates on the unit circle, I'm not sure whether that is the right approach.

Answer 1

Notice that the midpoint is: $$ M\left( \frac{c + a}{2}, \frac{d + b}{2} \right) $$

So the slopes of $OM$ and $PQ$ are, respectively: $$ \frac{d + b}{c + a} \qquad\text{and}\qquad \frac{d - b}{c - a} $$ Multiplying these together yields: $$ \frac{d^2 - b^2}{c^2 - a^2} $$ Now recall that $P$ and $Q$ are on the unit circle. So we may substitute both points into the equation of the unit circle to obtain: $$ a^2 + b^2 = 1 = c^2 + d^2 $$ Rearranging, notice that: $$ d^2 - b^2 = a^2 - c^2 = -(c^2 - a^2) \iff \frac{d^2 - b^2}{c^2 - a^2} = -1 $$ as desired.

Answer 2

If you're comfortable with vectors, they're the way to go - and we, as a nice bonus, incidentally prove the same thing about the sphere. Letting $\overrightarrow{P}$ and $\overrightarrow{Q}$ be the vectors to those points from the origin, we note that, being on the unit circle, their magnitude is $1$, and thus, using the dot product $$\overrightarrow{P}\cdot \overrightarrow{P}=\overrightarrow{Q}\cdot\overrightarrow{Q}=1$$ Further, note that $\overrightarrow{M}=\frac{1}2(\overrightarrow{P}+\overrightarrow{Q})$, being the midpoint, and is also a vector parallel the line $OM$. Moreover, a vector parallel the line $PQ$ is $\overrightarrow{P}-\overrightarrow{Q}$. For these to be perpendicular, their dot product must be zero: $$\frac{1}{2}(\overrightarrow{P}+\overrightarrow{Q})\cdot(\overrightarrow{P}-\overrightarrow{Q})=\frac{1}2(\overrightarrow{P}\cdot\overrightarrow{P}-\overrightarrow{Q}\cdot\overrightarrow{Q})=\frac{1}2(1-1)=0.$$ Thus, $OM$ and $PQ$ are orthogonal.