# The bisector CM of triangle ABC divides side AB into segments AM = 11 and MB = 16. The tangent to the circumscribed

**The bisector CM of triangle ABC divides side AB into segments AM = 11 and MB = 16. The tangent to the circumscribed circle of triangle ABC passing through point C intersects line AB at point D. Find CD.**

Consider the triangles ADC and CBD.

∠DCA = ∠CBA (since ∠DCA is equal to half the degree measure of the arc CA in the fourth property of the angles associated with the circle, and the inscribed angle CBA is also based on the same arc, which is also equal to half the degree measure of the arc, which is based on the theorem )

∠CDB is common to both triangles, therefore, by similarity, the triangles ADC and CBD are similar.

Therefore, by the definition of such triangles, we write:

CD / BD = AC / BC = AD / CD

AC / BC = AM / MB = 11/16 (according to the first property of the bisector).

From these equalities we write out:

AD = CD * 11/16

BD = CD * 16/11, (BD = AD + AB = AD + 16 + 11 = AD + 27)

AD + 27 = CD * 16/11

CD * 11/16 + 27 = CD * 16/11

27 = CD * 16/11-CD * 11/16

27 = (16 * 16 * CD-11 * 11 * CD) / 176

27 * 176 = CD (256-121)

CD = 4752/135 = 35.2

Answer: CD = 35.2