https://www.geogebra.org/geometry/xaj5mjuz
ORIGINAL QUESTION : You have given a circle with two diameters drawn. Select any point $P$ on the circle and drop perpendiculars to given diameters. The foot of perpendiculars are $X$ and $Y$. Prove that $XY$ is constant
The only thing that remains invariant is the two given diameters.After sometime playing with geogebra i noticed that we can consider the extreme case that is when $P$ coincide with $A$ or $B$ or $C$ or $D$.
Now we can rephrase the original question as
You have given a circle with two diameters drawn. Select any point $P$ on the circle and drop perpendiculars to given diameters. The foot of perpendiculars are $X$ and $Y$.Let $Q$ be the foot of perpendicular from point $A$ to $CD$.Prove that $XY$ is equal to $AQ$
Pure angle chasing does not solve the problem(I might have missed something).Normally olympiad geometry questions which are easy to state are generally more difficult.I think some kind of ingenious auxillary construction is required.If possible please provide your intuition and thought process along with solution.
The quadrilateral $PYOX$ is obviously cyclic. It follows $\angle PYX=\angle POX$. Continue the lines $(PY)$ and $(XO)$ till intersection at point $Z$. We have $$ \triangle ZYX\sim \triangle ZOP\implies \frac{XY}{OP}=\frac{ZY}{ZO}\implies XY=R\sin\theta, $$ where $R$ is the circle radius, and $\theta$ is the angle between the diameters.