A tangent AB is drawn through point C of a circle with center O, and AO = OB. Prove that AC = CB.

Let’s build the radius OC.

Since AB is tangent, point C is the point of tangency, and OC is the radius drawn to the point of tangency, then OC is perpendicular to AB, and then triangles OAC and OBC are rectangular.

By condition, ОВ = ОА, then in the right-angled triangles ОАС and ОВС the leg OC is common, and the hypotenuses of OA and OB are equal, therefore, the triangles are equal in the fourth sign – in the leg and hypotenuse.

Then AC = BC, as required.