Two chords AB and CD are drawn in the circle, intersecting at point E, AE = 12, CE = 8, DE-BE = 3. Find the product BE and DE.

Let the length of the segment BE = X cm, then DE – X = 3 cm.

DE = 3 + X cm.

Since the chords AB and СD 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: СE * DE = BE * AE.

8 * (3 + X) = X * 12.

24 + 8 * X = 12 * X.

4 * X = 24.

X = BE = 24/4 = 6 cm.

Then DE = 3 + 6 = 9 cm.

BE * DE = 6 * 9 = 54 cm2.

Answer: The product of BE and DE is 54 cm2.