Consider the following figure where $ORQP$ and $RSTQ$ are squares of same side length. $O,R,T$ lies on the circle with centre $B$.The radius of the circle is $5$ units. Find the side length of the squares.
My try:
Let the side of the squares be $a$.
Let $W$ be the mid point of $RT$. Then we know that $BW$ is $\perp$ to $RT$. Also $PR$ is $\perp$ to $RT$. Now i assumed that $RW \parallel PB$.
So we have $WB=PR=a\sqrt{2}$
By Pythagoras theorem in $\Delta BRW$ we get $$25=(a\sqrt{2})^2+\left(\frac{a}{\sqrt{2}}\right)^2$$
Which gives $a^2=10$ and hence $a=\sqrt{10}$.
But geometrically how to prove that $RW \parallel PB$?
Take quadrant formed by $\angle ABU$.
Firstly, $QR = QT$ and $\angle RQT = 90^0$ so it must be on the perpendicular bisector of $RT$, which is $BW$.
That gives us $\angle TQW = \angle BQP = \angle BPQ = 45^0$. We also have $\angle RPQ = 45^0$.
So $\angle RPB = 90^0$ and $PR \parallel BW$.
Also see the $9$ squares that Calvin Lin referred to in comments.
That leads to side of the square $s$ given by,
$(3s)^2 + s^2 = d^2 \implies s = \frac{d}{\sqrt{10}}$
where $d$ is the diameter of the circle.