Prove the tangent secant segments theorem: If a secant through an external point P meets a circle at B and C (P – B – C), and PA is a tangent segment from P touching the circle at A, then \(PA^2 = PB \times PC\).

Answer: \(PA^2 = PB \times PC\) (proved below)

[Figure: circle; external point P; tangent PA touching at A; secant P – B – C; seg AB and seg AC drawn.]

Construction: Draw seg AB and seg AC.

Proof: \(\angle PAB = \dfrac{1}{2} m(\text{arc AB})\) (tangent secant angle theorem) and \(\angle ACB = \dfrac{1}{2} m(\text{arc AB})\) (inscribed angle theorem). \(\therefore \angle PAB \cong \angle PCA\).

In \(\Delta PAB\) and \(\Delta PCA\): \(\angle APB \cong \angle CPA\) (common angle) and \(\angle PAB \cong \angle PCA\) (shown above).

\(\therefore \Delta PAB \sim \Delta PCA\) (AA test) \(\therefore \dfrac{PA}{PC} = \dfrac{PB}{PA}\) (corresponding sides of similar triangles).

\(\therefore PA^2 = PB \times PC\). Hence proved.