TheoremProved

Power of a Point (Intersecting Chords and Secants)

Statement

If two lines through a point $P$ intersect a circle at the endpoints of chords $AB$ and $CD$ , then the products of the segment lengths from that point are equal: $PA \cdot PB = PC \cdot PD$ . When the point lies outside the circle and a tangent segment to the circle is drawn, its squared length satisfies $PT^2 = PA \cdot PB$ .

Why is it true?

Although a circle is curved, the inscribed angle theorem locks the angles of the two triangles formed by the intersecting chords into equality, making the triangles similar; converting their side-length proportion into a cross-product turns a ratio of lengths into an invariant product.

Proof sketch

Step 1 (form two triangles using the chord endpoints). Join the endpoints by segments $AC$ and $BD$ to form the two triangles $\triangle PAC$ and $\triangle PDB$ .

Step 2 (prove the two triangles are similar via inscribed angles). Because inscribed angles subtending the same arc are equal, we have $\angle CAP = \angle BDP$ . In addition, the angles at the intersection point are equal (as vertical angles when the point is inside the circle, or as the same shared angle when the point is outside): $\angle APC = \angle DPB$ . By the Angle-Angle similarity criterion, $\triangle PAC \sim \triangle PDB$ .

Step 3 (take the ratio of corresponding sides and cross-multiply). Similarity of the two triangles gives the side ratio $\dfrac{PA}{PD} = \dfrac{PC}{PB}$ . Cross-multiplying both sides yields $PA \cdot PB = PC \cdot PD$ .

Topics that use this theorem

Circles

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

Euclid (trans. Thomas L. Heath) (1956). Euclid's Elements (Books I–XIII)
H. S. M. Coxeter, S. L. Greitzer (1967). Geometry Revisited