Perpendiculars drawn to the sides of the angle AOB at points A and B intersect at point C

Perpendiculars drawn to the sides of the angle AOB at points A and B intersect at point C lying inside this angle, prove that a circle can be described around the quadrilateral ACBO.

Let’s construct an additional OS segment.

Two right-angled triangles are formed: OAC and OBC with right angles B and A (by construction).

These triangles have a common OS hypotenuse.

A circle can be described around a right-angled triangle, the center is at the point dividing the hypotenuse in half. The radius of the circle r will be equal to half the hypotenuse.

The hypotenuse of the OS is common for the triangles, so the radii of the circles are equal (r = OC / 2).

The centers of the circles coincide and are located at point M.

From this it follows that the circles around the two triangles also coincide. In fact, this is one circle passing along the vertices of the AOVS quadrilateral.