The diameter of the circle intersects the chord AB at point C and is perpendicular to it. prove that AC = CB.

Let’s construct the radii ОА and ОВ. The AOB triangle is isosceles, since ОА = ОВ = R.

By condition, the chord AB is perpendicular to the diameter of the CD, then OH is the height, median and bisector of the triangle AOB. AH = BH.

Triangles AON and BON are rectangular and equal in leg and hypotenuse, and then the angle COA = COB, which means that the arcs AC and BC are also equal, and therefore the chord AC = AB, which was required to be proved.