Diameter AB intersects the chord CB at point M. Find the segments into which point M divides

Diameter AB intersects the chord CB at point M. Find the segments into which point M divides the diameter AB, if r = 10 cm, CM = 4 cm, MD = 9 cm.

Let the length of the segment OM = X cm, then the segment AM = R – X = 10 – X cm, the length of the segment CM = R + X = 10 + X cm.

AB and CD are intersecting chords, by the property of which the product of the lengths of the segments into which they are divided at the intersection point are equal.

AM * CM = CM * DM.

(10 – X) * (10 + X) = 4 * 9.

100 – 10 * X + 10 * X – X ^ 2 = 36.

X ^ 2 = 64.

X = 8 cm.

Then AM = 10 – 8 = 2 cm, CM = 10 + 8 = 18 cm.

Answer: Point M divides the diameter AB into segments of 2 cm and 18 cm.