How to Prove Points are Cyclic and the Power of a Point

Proving Points are Cocyclic

To prove that points \( A \), \( B \), \( C \), and \( D \) are cocyclic, we can use the inscribed angle theorem, the central angle theorem, the power of a point theorem, or the exterior angle criterion.

\[ \text{If } \angle ABC = \angle ADC, \text{ then } A, B, C, D \text{ are cocyclic.} \]

\[ \text{If } PA \times PC = PB \times PD, \text{ then } A, B, C, D \text{ are cocyclic.} \]

Illustrative Diagram

In this diagram, points \( A \), \( B \), \( C \), and \( D \) are shown to be on the same circle, proving they are cyclic.

Power of a Point

The power of a point \( P \) relative to a circle with center \( O \) and radius \( r \) is defined as the squared distance from \( P \) to \( O \), minus the square of the radius:

\[ \text{Power of } P = PO^2 - r^2 \]

If \( P \) lies on the circle, the power of \( P \) is zero. Otherwise, the power is positive if \( P \) is outside the circle and negative if \( P \) is inside.

Exercise 1: Proving Points are Cocyclic

Given four points \( A \), \( B \), \( C \), and \( D \) in a plane, prove that these points are cocyclic if the following condition holds:

\[ \angle BAC + \angle BDC = 180^\circ \]

Solution to Exercise 1

By the inscribed angle theorem, \( \angle BAC + \angle BDC = 180^\circ \) implies that the points \( A \), \( B \), \( C \), and \( D \) lie on the same circle. Hence, they are cocyclic.

Exercise 2: Power of a Point

Given a circle with center \( O \) and a point \( P \) outside the circle, show that the power of \( P \) with respect to the circle equals the product of the lengths of the tangents from \( P \) to the circle.

Solution to Exercise 2

The power of \( P \) is given by:

\[ \text{Power of } P = PT_1 \times PT_2 \]

where \( T_1 \) and \( T_2 \) are the points of tangency. This shows that the power of \( P \) is equal to the product of the lengths of the tangents from \( P \) to the circle.