Two congruent chords AB and CD which are not diameters, intersect at right angle in P. O is the centre of the circle. If M and N are the mid-points of AB and CD respectively, then prove that quadrilateral OMPN is a square.
If horizontal chord is AB and vertical chord is CD, then it will be a square for figure OMPN.
From what I've observe, the diameter will be √2 times the leg.
From this perspective, midpoint of each leg is 0.5 or 1/2 of original length (1 or x). And midpoint of diameter (hypotenuse) will be √2/2. From here, you have to calculate the new lengths of each leg to see if they add up to half of diameter (hypotenuse) (a2 + b2 = c2) and to see if each leg are perpendicular or form a right angle. Basically, you will have a square inside of 45-45-90 triangle.