1. The incircle of triangle ABC touch sides AB,BC, CA at M,N,K, respectively. The line through A parallel to NK meets MN at D. The line through A parallel to MN meets NK at E. Show that the line DE bisects sides AB and AC of triangle ABC.
Solution: We may assume angle A > angle C ( equivalently a > c) Extend AD, AE to meet the sideline BC at D' and E' respectively. CM = CN and MN // D'A imply CD' = CA = b BM = BK and KM // AE' imply BE' = BA = c Consider the intouch triangle KMN. Denote its angles by K, M, N I is its circumcircle So MN = 2r Sin K = 2r Sin (90 - C/2) = 2r Cos C/2 = 8R SinA/2 Sin B/2 SinC/2 Cos C/2 Next we find EN by applying Sine Rule to Triangle AEN First note AN = s - a, angle AEN = M = 90 - A/2 Angle EAN = A - angle BAE'= A -(180 - B)/2 =(A - C)/2 Now EN / Sin EAN = AN / Sin AEN So EN = [(s - a )Sin(A - C)/2]/Cos A/2 = [(rCot A/2)Sin(A - C)/2]/Cos A/2 = [r Sin(A - C)/2]/Sin A/2 = [4R SinA/2 Sin B/2 SinC/2 Sin(A - C)/2]/Sin A/2 = 4R Sin B/2 SinC/2 Sin(A - C)/2 Follows AD = EM = MN - EN = (4R Sin B/2 SinC/2)[2 Sin A/2 Cos C/2 - Sin(A - C)/2] = 4R Sin B/2 SinC/2 Sin (A/2 + C/2) = 4R Sin B/2 SinC/2 Cos B/2 = 2R Sin B Sin C/2 = b Sin C/2 = (1/2)AD' (from the isosceles triangle AD'C in which AD' = AC = b and vertical angle = C) Hence D is the midpoint of AD' By a similar argument E is the midpoint of AE' Thus in triangle AD'E', DE is the join of midpoints of the sides AD', AE' Follows that the line DE is parallel to BC Observe that ADME being a parallelogram DE bisects AM say at X. Let the line DE cut AB at Z and AC at Y. Now in triangle ABM, X is the midpoint of AM and YX // BM Therefore Y is the midpoint of AB and similarly Z is the midpoint of AC, thereby proving that the line DE bisects both AB and AC Q.E.D