Question 41
Let $AB$ be a diameter of a circle with centre $O$. We choose a point $C$ on the circumference of the circle such that $OC$ and $AB$ are perpendicular to each other. Let $P$ be an arbitrary point on the (smaller) arc $BC$ and let the lines $CP$ and $AB$ meet at $Q$. We choose $R$ on $AP$ so that $RQ$ and $AB$ are perpendicular to each other.
Show that $|BQ| = |QR|$.
Show that $|BQ| = |QR|$.