Motion down an inclined plane with leg equal to the diameter of a circle This is an exercise from Morris Kline's "Calculus: An Intuitive and Physical Approach".

An object slides down an inclined plan $OP'$ (Fig. 3-9) starting from rest at $0$. Show that the point $Q$ reached in the time $t_1$ required to fall straight down to $P$ lies on a circle with $OP$ as diameter.


The text gives a proof by contradiction for this problem:

From the formula $s = 16t^2$ we find that the time to fall the distance $OP$ is $t_1 = \sqrt{OP}/4$. For the motion along $OP'$ we use (35), that is, $s = 16t^2 \sin{A}$. The distance $OQ$ that the object slides in time $t_1$ is $OQ = 16(OP/16) \sin{A}$. Then $ \sin{A} = OQ / OP$. Suppose $Q$ is not on the circle but R on $OP'$ is. Then $\angle OPR$ is $ \angle A$ by the use of right triangles. Then $\sin A = OR/OP$. But $\sin{A} = OQ/OP$. Hence $Q = R$ and $Q$ lies on the circle.

This proof is confusing to me because I do not know the logical form of the statement we are trying to prove. What if $Q$ does not intersect with the circle at all? What is the justification for the bold portions of the proof?
Is it possible to directly show that $Q$ lies on the circle?
My approach is to let $C$ be the center of the circle. If we can show that the length of $CO$ is equal to the length of $CQ$, then $Q$ will be on the circle.
Although, I keep getting stuck trying to show that $CQ = \frac{OP}{2}$.

For both paths.
Time = $\frac 12 \frac {\text {distance}}{\text {acceleration}}$
On the trip from $O$ to $Q,$ the acceleration is $g\sin A$
The distance covered is $\overline {OP} \sin A$
The alternative trip from $O$ to $P$ the accelartion is $g$ and the distance is $\overline {OP}$
