1. **Problem Statement:** Given two chords AB and CD of a circle intersecting at point E inside the circle, prove that $|AE| \cdot |EB| = |CE| \cdot |ED|$. 2. **Key Idea:** Use the similarity of triangles formed by the intersecting chords. 3. **Step 1: Identify triangles** The chords intersect at E, creating two pairs of triangles: $\triangle AEC$ and $\triangle DEB$. 4. **Step 2: Show similarity** Since A, B, C, D lie on the circle, angles subtended by the same chord are equal: - $\angle AEC = \angle DEB$ (vertical angles) - $\angle CAE = \angle BDE$ (angles subtended by chord CD) Thus, $\triangle AEC \sim \triangle DEB$ by AA similarity criterion. 5. **Step 3: Write ratio of corresponding sides** From similarity, corresponding sides are proportional: $$\frac{|AE|}{|DE|} = \frac{|CE|}{|BE|} = \frac{|AC|}{|DB|}$$ 6. **Step 4: Cross multiply to get the required relation** $$|AE| \cdot |BE| = |CE| \cdot |DE|$$ 7. **Conclusion:** We have proved that the product of the segments of one chord equals the product of the segments of the other chord when two chords intersect inside a circle.