Théorème des cordes sécantes

Problem

Two chords cross inside a circle: $AP = 4$, $PB = 9$, and $CP = 6$. Find $PD$, and if $AB$ is a diameter, find the radius.

Step 1: Use the intersecting chords theorem

For two chords that intersect at $P$, the segment products are equal:

$$AP \cdot PB = CP \cdot PD$$

Substitute the given values:

$$4 \cdot 9 = 6 \cdot PD$$

So,

$$36 = 6PD$$

Dividing by $6$ gives

$$PD = 6$$

Step 2: Find the diameter and radius

Since $AB$ is a diameter, its length is the sum of the two chord segments:

$$AB = AP + PB = 4 + 9 = 13$$

The radius is half the diameter:

$$r = \dfrac{13}{2} = 6.5$$

Answer

$$PD = 6 \quad \text{and} \quad r = 6.5$$