Slope of Tangent to Cycloid/Proof 2

Proof
Consider a polygon $ABCD$ being rolled along a straight line in the same way as the generating circle of $C$.

Let $A', B', C', D'$ be the points around which the $ABCD$ rotates while rolling.


 * TangentToCycloid-construction.png

The point $A$ traces out in succession several arcs of circles with centers $B', C', D'$.

The tangent to each of these arcs is perpendicular to the line joining the point of tangency to the corresponding point of rotation.

Consider a circle as being a polygon with an infinite number of sides.

It follows that in the limit the tangent to $C$ at any given point is perpendicular to the line joining the point of tangency to the corresponding point where the generating circle touches the straight line along which the generating circle rolls.


 * TangentToCycloid.png

Let $O$ be the center of the generating circle.

Let $P$ be the point of tangency.

Let $Q$ be the point where the generating circle touches the straight line along which the generating circle rolls.

Let $S$ be the point where the tangent to $C$ meets the straight line along which the generating circle rolls.

As $OP$ and $OQ$ are both radii of the generating circle:
 * $OP = OQ = a$

and so $\triangle POQ$ is isosceles.

Let $R$ be dropped perpendicular to $PQ$ at $R$.

As $\triangle POQ$ is isosceles, $\triangle OPR = \triangle OQR$ and so $PR = RQ$.

Thus $\angle POR = \angle QOR = \dfrac \theta 2$.

As $OR \perp PR$ and $PR \perp PS$, we have that $\triangle OPR$ is similar to $\triangle QPS$.

Thus $\angle PQS = \angle POR$ and so $\angle PSQ$ is $\dfrac \pi 2 - \angle POR = \dfrac \pi 2 - \dfrac \theta 2$.

By definition, $\tan \angle PSQ$ is the slope of the tangent $\dfrac {\mathrm d y} {\mathrm d x}$ to $C$.

Thus:

Historical Note
This proof was discovered by.