```Date: Aug 29, 2006 3:04 AM
Author: Vijaya Prasad Nalluri
Subject: Re: Help in Geometry

1. The incircle of triangle ABC touch sides AB,BC, CA at M,N,K, respectively. The line through Aparallel to NK meets MN at D. The line through A parallel to MN meets NK at E. Show that theline 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 = bBM = BK and KM // AE' imply BE' = BA = cConsider the intouch triangle KMN. Denote its angles by K, M, NI 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/2Next we find EN by applying Sine Rule to Triangle AENFirst note AN = s - a, angle AEN = M = 90 - A/2Angle EAN = A - angle BAE'= A -(180 - B)/2 =(A - C)/2Now EN / Sin EAN = AN / Sin AENSo 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)/2FollowsAD = 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 BCObserve that ADME being a parallelogram DE bisects AMsay 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 // BMTherefore 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
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