# Two chords are drawn in the circle: CD and AB intersect at point E

**Two chords are drawn in the circle: CD and AB intersect at point E, so that AE is 4 cm more than BE, DE is 16 cm less than CE. CE: DE = 3: 1. Find AE, BE, CE, DE**

By condition, CE – DE = 16 cm, and CE / DE = 3/1.

Let’s solve the system of equations.

CE = 3 * DE.

Then 3 * DE – DE = 16.

DE = 16/2 = 8 cm.

CE = 3 * 8 = 24 cm.

Let the length of the segment AE = X cm, then the length of the segment BE = (X – 4) cm.

Since the chords AB and CD intersect at the point E, then by the property of intersecting chords, the product of the lengths of the segments obtained at their intersection, of one chord, is equal to the product of the segments of the other chord.

Then: CE * DE = BE * AE.

24 * 8 = (X – 4) * X.

192 = X ^ 2 – 4 * X.

X ^ 2 – 4 * X – 192 = 0.

Let’s solve the quadratic equation.

X = AE = 16 cm.

Then BE = 16 – 4 = 12 cm.

Answer: AE = 16 cm, BE = 12 cm, CE = 2 cm, DE = 8 cm.