AB and CD are two equal chords of a circle with centre O , that subtend a right angle at X . P and Q are the midpoints of AB and CD respectively . Prove that OPXQ is a square .
Hello ukorovIf your diagram is correct, then the question is not misleading - it's just plain wrong. Nothing is subtending an angle of $\displaystyle 90^o$ at $\displaystyle X$. And how do you know that $\displaystyle X$ is the point of intersection of the chords? They are described as $\displaystyle AB$ and $\displaystyle CD$, not $\displaystyle AX$ and $\displaystyle DX$.
Grandad