A quadrilateral with mutually perpendicular diagonals is inscribed in a circle with center O.

A quadrilateral with mutually perpendicular diagonals is inscribed in a circle with center O. Prove that the distance from point O to each of its sides is equal to half the length of the opposite side.

Consider a quadrilateral ABCD inscribed in a circle centered at O, for which

diagonals AC and BD are perpendicular.

Draw a straight line through points D and O. Let the line DO intersect the circle at point E.

Obviously DE is the diameter of the circle. Therefore, the triangle DBE is right-angled and the angle DBE is right. This implies that lines AC and BE are parallel, since both lines are perpendicular to line BD.

Since AC || BE, then angle CAB = ABE. Consequently, the arcs AE and BC are equal, and hence the contracting chords AE = BC are also equal.

Note that the angles are CAE = ACB. Therefore, triangles ACE and ACB are equal in two sides and the angle between them. This implies that AB = CE.

Since the chords AB and CE are equal, the distances to them from the center of the circle are also equal:

OG = OH.

Since OE = OC, H is the middle of CE. Hence,

OH is the middle line of the triangle ECD and

OH = CD / 2.

So we got that OG = OH = CD / 2, which is what was required to prove.