In triangle ABC, angle A is 40 degrees, angle B = 20 degrees, and AB-BC = 4. Find the length of the bisector of angle C.

Let CK be the bisector of the angle C.

Let us postpone point D on AB so that BD = CB.

Then the angle BDC = BCD = (180º – 20º) / 2 = 80º.

ACB angle = 180º – (40º + 20º) = 120º.

Then the angle ACK = 60º, since CK is a bisector.

AKC = 180º – (ACK + CAK) = 180º – (60º + 40º) = 80º.

Consider triangle DCK – isosceles, since the angle CDK = CKD = 80º.

Therefore, CD = CK, angle DCK = 180º – 2 * 80º = 20º.

Then the angle ACD = ACK – DCK = 60º – 20º = 40º.

Then triangle CDA is isosceles, since the angle CAD = ACD = 40º.

And since AD = AB – DB = AB – BC = 4, then CD = AD = 4.

But CD = CK, hence CK = 4.