The chords AB and CD of the circle intersect at point E. AE refers to EB as 1 to 3. CD = 20. BE = 5. Find AB.

Consider triangles AEC and BED in which the angles at the vertex E are equal as vertical angles. The inscribed angles ACD and ABD are based on one arc AD, then these angles are equal, and therefore the triangles AEC and BED are similar in two angles.

Then ME / KE = PE / NE.

Let the length of the segment AE = X cm, then BE, by condition, is equal to 3 * X cm.

CE = CD – DE = 20 – 5 = 15 cm.

AE / CE = DE / BE.

X / 15 = 5/3 * X.

3 * X ^ 2 = 5 * 15 = 75.

X ^ 2 = 25.

X = AE = 5 cm.

BE = 3 * 5 = 15 cm.

AB = AE + BE = 5 + 15 = 20 cm.

Answer: The length of the chord AB is 20 cm.