Prove that the bisectors of the outer and inner angles of a triangle drawn from one vertex are perpendicular.

Let the inner angle ACB = X0.

The external angle BCH is adjacent to the angle ACB, the sum of which is 180, then the angle BCH = (180 – X) 0.

Since the CК is a bisector, the angle KCB = ACB / 2 = (X / 2) 0.

Similarly, the angle ВСМ = ВСН / 2 = (180 – X) / 2 = (90 – X / 2) 0.

Then the angle KСM = KСB + BCM = X / 2 + 90 – X / 2 = 90, therefore KS is perpendicular to CM, one hundred and it was required to prove.