# Geometrical problem in Newton's “Principia”.

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?

• $\triangle ZQR\sim \triangle ZTP$ because $\overline{QR}\parallel\overline{TP}$. For the other similarity, first note that $\angle VPA$ is a right angle (via Thales' Theorem). Also, if we introduce $O$ as the center of the circle (aka, the midpoint of $\overline{VA}$), we can do a little angle-chasing to show $$\angle ZPT=90^\circ−\angle OPT=\angle OPA = \angle A$$ Thus, $\triangle ZTP\sim\triangle VPA$ by Angle-Angle Similarity. – Blue Jan 6 at 18:44

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$$\triangle ZTP\sim\triangle VPA$$