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Topic: Help in Geometry
Replies: 3   Last Post: Aug 29, 2006 3:04 AM

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 Vijaya Prasad Nalluri Posts: 8 From: India Registered: 4/10/06
Re: Help in Geometry
Posted: Aug 29, 2006 3:04 AM

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
AD' = AC = b and vertical angle = C)
Hence D is the midpoint of AD'
By a similar argument E is the midpoint of AE'
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

Date Subject Author
7/12/06 Wong wane young
8/16/06 Alexander Bogomolny
8/16/06 Alexander Bogomolny