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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 $PV^2$.enter image description here

My question is how can we prove the similarity of triangles ZQR, ZTP, and VPA?

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    $\begingroup$ $\triangle ZQR\sim \triangle ZTP$ 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. $\endgroup$ – Blue Jan 6 at 18:44
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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$$\triangle ZTP\sim\triangle VPA$$

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