We are given the circle with center at *O*, and a point *A* outside the circle. Connect *AO*, and let *M* be its midpoint.

Draw the circle centered at *M* going through *O*. Since *M* is the midpoint of *AO*, it will also go through *A*. Let *B* and *C* be the points where this circle intersects the given circle. Then *AB* and *AC* will be the lines through *A* tangent to the given circcle.

This is because __/__*ABO* and __/__*ACO* will be inscribed angles in the circle centered at *M*, so their measure will be half of the central angle. But since *AMO* is a diameter of the circle, the central angle is a straight angle, so __/__*ABO* and __/__*ACO* will be right angles. Since the tangent is perpendicular to the radius at the point of tangency, *AB* and *AC* will be tangents to the circle.

This construction will help us use the Isosceles Triangle Theorems to prove a theorem about tangents.

Theorem: Given a circle and a point outside the circle, the line from the point to the center of the circle will bisect the angle formed by the tangents, and the point is equidistant from the two points of tangency.

Proof:* BC* will be perpendicular to *AO* because the line that joins the two points where two circles intersect will be perpendicular to the line that joins their centers. *MB* and *MC* will be congruent, because they are both radii of the same circle. Therefore triangle *MBC* is isosceles, and the line that is perpendicular to the base will bisect the vertex angle. Thus, __/__*BAO* and __/__*CAO *which are inscribed angles for these congruent angles will also be congruent, and *AO* will bisect the angle formed by the tangents.

In triangle *ABC* the line which bisects the vertex angle is perpendicular to the base, so the triangle is isosceles.

It is possible to prove these results simply using congruent triangles, but this construction enables us to use the Isosceles Triangle Theorems.