Geometry Problem

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## Wednesday, March 2, 2011

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

Labels:
half difference,
incenter,
incircle,
midpoint,
tangency point,
tangent,
triangle

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Name E, F tgpoints on AB, BC

ReplyDeleteAD+x = DC-x => 2x = DC-AD = CF-AE = (CF+BF)-(AE+BE)

2x = BC-AB = a - c

x = (a - c)/2

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

ReplyDeleteThen

2MD

=CD-AD

=CF-AE

=(CF+BF)-(AE+BE)

=BC-AB

=a-c

i shall use the usual notation for the sides of the triangle and the semi perimeter.

ReplyDeleteAB = 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.