The chord CD of the circle centered at point O is perpendicular to the radius OA

The chord CD of the circle centered at point O is perpendicular to the radius OA and passes through its midpoint F. Prove that the quadrilateral is ACOD-rhombus.

By condition, the chord СD divides the radius OA in half, then AF = OF.

Let us draw the radii OS and OD from the center O circle to the edges of the chord, then the triangle AOD will be isosceles, OS = OD = R.

The height OF in the СOD triangle is also its median, then CF = DF = СD / 2.

The height CF is perpendicular to the СD and lies on the radius OA, then OA is perpendicular to the СD.

In the AСOD quadrangle, the diagonals OA and СD are perpendicular, and at the point F they are divided in half, therefore, the AСOD quadrilateral is a rhombus, which was required to prove.