This theorem states that if we draw two lines from the same point which lies outside a circle, such that both lines are tangent to the circle, then their lengths are the same. We will now prove that theorem.

## Problem

AB and AC are tangent to circle O. Show that AB=AC

## Strategy

To show two lines are equal, a helpful tool is triangle congruency. The shape of the drawing in the problem statement - where the two tangent lines create a sort of triangular 'clown's hat' also suggests we would be served by constructing some triangles here, where the two tangent lines are sides.

We also know that a property of the tangent lines to a circle is that they form a 90° angle between the line and a radius at the point of tangency - so let's draw that:

And now the triangles almost present themselves. Connect the point A with a line segment to the center of the circle, O, and we will have two right triangles, with a common hypotenuse (AO), and an equal leg, as both radii are equal, and the triangles are congruent by HL.

## Proof

(1) AB is tangent to Circle O //Given

(2) ∠ABO=90° //tangent line is perpendicular to circle

(3) AC is tangent to Circle O //Given

(4) ∠ACO=90° //tangent line is perpendicular to circle

(5) AO=AO //common side (reflexive property)

(6) OC=OB=r //radii of a circle are all equal

(7) △ABO≅△ACO //Hypotenuse-leg

(8) AB=AC // Corresponding sides in congruent triangles (CPCTC)

This theorem is a key element in proving the Pitot Theorem, about a quadrilateral that circumscribes a circle.