Question
Prove that the lengths of tangents drawn from an external point to a circle are equal.
That is, if PA and PB are tangents from external point P to a circle with centre O, prove that PA = PB.
Solution — Step by Step
Join O to P, O to A, and O to B. We now have two triangles: △OAP and △OBP. The plan is to prove these triangles congruent — then PA = PB falls out directly.
Since PA is a tangent at A, and OA is the radius to that point, we know OA ⊥ PA. So ∠OAP = 90°. By the same logic, OB ⊥ PB, giving ∠OBP = 90°. This is the key property: radius to point of tangency is always perpendicular to the tangent.
In △OAP and △OBP:
- OA = OB (radii of the same circle)
- OP = OP (common side)
- ∠OAP = ∠OBP = 90°
We have a Right angle, the Hypotenuse OP is common, and one Side OA = OB. So by RHS congruence, △OAP ≅ △OBP.
Since the triangles are congruent, their corresponding parts are equal. Therefore PA = PB. ∎
Why This Works
The whole proof rests on one circle theorem: a radius drawn to a point of tangency is perpendicular to the tangent. Without that 90°, we’d have no right triangles, and RHS wouldn’t apply at all.
Once we have two right triangles sharing a hypotenuse and with equal shorter sides, RHS guarantees they’re mirror images of each other. PA and PB are corresponding sides in those mirror-image triangles — so they must be equal.
This result is used constantly in coordinate geometry and mensuration problems. When a circle is inscribed in a triangle, we use this fact at every vertex to find the tangent lengths — it’s the engine behind those “find the sides” questions.
Alternative Method (Using Pythagoras)
If you’re more comfortable with Pythagoras than congruence, here’s another route.
In △OAP: , so
In △OBP: , so
Since OA = OB (radii), the right-hand sides are equal. So , which gives PA = PB.
In a 3-mark proof question, the RHS congruence method is what CBSE expects and awards full marks for. The Pythagoras method is great for understanding but write it only if the question specifically says “using Pythagoras” or as supplementary work.
Common Mistake
Students write “OP = OP (common)” and think that’s enough to claim congruence — then pick SAS or SSS without checking all conditions. RHS specifically needs: one angle = 90°, hypotenuses equal, one pair of sides equal. Skipping the explicit statement “∠OAP = ∠OBP = 90° because radius ⊥ tangent” will cost you marks. CBSE board checkers look for this line.
A second trap: some students try to prove the triangles congruent using ASS (two sides and a non-included angle) — that’s not a valid congruence rule. The reason RHS works here is that the 90° angle is specifically at the side, not somewhere in the middle of the triangle. We’re in safe territory.