Two perpendicular chords AC and BD are drawn in the circle, intersecting at point M

Two perpendicular chords AC and BD are drawn in the circle, intersecting at point M. Prove that the line passing through M perpendicular to AB divides CD in half.

The KM segment is the height of a right-angled triangle drawn to the hypotenuse.

Then the triangles ABM, KВM and AKM are similar.

Angle BAM = ВMK, angle ABM = AMK.

Angle BAM = BAC = BDC since they are based on one arc BC.

The angle DМН = ВМК as vertical angles, then the angle DМК = МDС = МDК, and then the triangle МDН is isosceles, МН = DH.

Angle ABM = ABD = ACD as they rest on the arc AD.

Angle АМК = СМН as vertical angles.

Then the angle СВН = MCH, which means that the triangle MCH is isosceles, MH = CH.

Since MH = CH = DH, then point H is the middle of CD, and then CH = DH, which was required to prove.