Proving equal tangent lengths

If PA and PB are tangents to a circle with centre O, the radii OA and OB meet the tangents at right angles. The point P must lie outside the circle and A and B must be the points of contact. The triangles OAP and OBP share the hypotenuse OP and have equal radii OA and OB. The triangles are congruent by RHS: right angle, hypotenuse and side.

Write the correspondence between the triangles before concluding that particular sides or angles are equal. Here A corresponds to B, while O and P correspond to themselves. A diagram suggests the result but does not supply the radius-tangent theorem or the congruence argument. An alternative proof applies Pythagoras in both triangles and subtracts the equal squared radii from the same squared hypotenuse.

Worked example

PA and PB touch a circle with centre O at A and B. Prove that PA = PB.

1. OA is perpendicular to PA and OB is perpendicular to PB, by the radius-tangent theorem.
2. OA = OB because both are radii, and OP is common to the two right-angled triangles.
3. Triangles OAP and OBP are congruent by right angle, hypotenuse and side. Corresponding tangent sides are equal.

Answer: PA = PB.

Revise first: Circle theorems and angle reasoning.