Wednesday, March 2, 2011

Problem 587: Triangle, Incenter, Incircle, Tangency Point, Midpoint, Distance, Half the Difference

Geometry Problem
Click the figure below to see the complete problem 587.

 Problem 587: Triangle, Incenter, Incircle, Tangency Point, Midpoint, Distance, Half the Difference.
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3 comments:

  1. Name E, F tgpoints on AB, BC
    AD+x = DC-x => 2x = DC-AD = CF-AE = (CF+BF)-(AE+BE)
    2x = BC-AB = a - c
    x = (a - c)/2

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  2. E,F tangency points of AB, AC with incircle, say.

    Then

    2MD
    =CD-AD
    =CF-AE
    =(CF+BF)-(AE+BE)
    =BC-AB
    =a-c

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  3. i shall use the usual notation for the sides of the triangle and the semi perimeter.
    AB = c, BC = a and AC = b and s = (1/2)*(a+b+c)
    now by the tangency property of the in circle we have AD = s-a and AM = b/2 as M is the midpoint of AC. now DM = AM - AD = b/2 - (s-a) = a/2 - c/2 = (1/2)*(a-c). this implies that DM = (1/2)*(BC - AB).
    Q. E. D.

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