Let VQPA be the circumference
of the circle, S the given
point toward which the force
tends as to its center, P the body
revolving in the circumference, Q
the place to which it will move
next, and PRZ the tangent of
the circle at the previous place.
Through point S draw chord PV;
and when the diameter VA of
the circle has been drawn, join
AP; and to SP drop perpendicular
QT, which when produced meets the tangent PR at Z; and finally through
point Q draw LR parallel to SP and meeting both the circle at L and the
tangent PZ at R. Then because the triangles ZQR, ZTP, and VPA are
similar, $RP^2$ (that is, QR x RL) will be to $QT^2$ as $AV^2$ to $PV^2$.
My question is how can we prove the similarity of triangles ZQR, ZTP, and VPA?
Accepting the comment of @Blue on the similarity of triangles $ZQR$ and $ZTP$, then without further construction, since $PZ$ is tangent at $P$, in right triangles $ZTP$ and $VPA$, $\angle ZPT=\angle VAP$ (Euclid, Elements III, 32). Therefore$$\triangle ZTP\sim\triangle VPA$$