Let ABC be a triangle with sides of lengths a, b and c. Let the bisector of the angle C cut AB in D.

Prove that the length of CD is 2ab·cos(C/2)/(a + b).

**Solution by Steve Dinh, a.k.a. Vo Duc Dien**

Extend BC to the right a length of CE = AC = b. From A draw the perpendicular to AE to meet the extension of CB to the left at F.

Since AC = CE, ∠AEB = ½ ∠ACB = ∠DCB, and CD || AE and we have

CD/AE = a/(a + b), or CD = a·AE /(a + b) (1)

We also have ∠AFE = 90°- ∠AEF = 90°- ∠CAE = ∠FAC,

or CAF is isosceles with CA = CF = b,

and cos(C/2)=cos∠AEF = AE/(2b),

or AE = 2b·cos(C/2), and (1) now becomes

CD = 2ab ·cos(C/2)/(a + b)