The following is a standard fact about circles:
THEOREM: Let $p$ and $q$ be two antipodal points on a circle in $\mathbb{R}^2$ and let $r$ be another point on the circle such that $r \neq p,q$. Then the angle formed by the line segments $pr$ and $rq$ is a right angle.
It is easy enough to prove this theorem using cartesian coordinates, but I'm looking for synthetic proofs (as many as possible). Thanks!
On
The simplest is to consider the triangles $orp$ and $orq$. They are both isosceles because $or=op=oq$ as radius of the circle. Therefore the angles at their base are equal. Which means $\widehat{opr}=\widehat{orp}$ and $\widehat{oqr}=\widehat{orq}$. Besides in the triangle $prq$ one has
$$\widehat{pqr}+\widehat{qpr}+\widehat{prq}=\pi$$
and
$$\widehat{prq}=\widehat{orp}+\widehat{orq}=\widehat{qpr}+\widehat{pqr}$$
From those two equations we get $\widehat{prq}=\frac{\pi}{2}$
On
Here is another argument. The statement of the theorem is suggesting that the answer does not depend on the choice of $r$ nor the choice of antipodal points $p$ and $q$. So assume this is indeed the case and let call $\alpha$ the angle between $pr$ and $rq$. Let $r'$ be the antipodal of $r$. Now, for instance, the angle between $rq$ and $qr'$ is again $\alpha$ because $r,r'$ are antipodal and $q \neq r,r'$. The same for the two other angles. So you get $4\cdot \alpha = 360^{\circ}$ and $\alpha = 90^{\circ}$.
The inscribed angle formed intercepts 180$^{\circ}$ of arc. So it's measure is $90^{\circ}.$See this.