Let ABC be a triangle with |AB| > |AC|. Let D be a point on line AB such that |AD| + |DC| = |AB| and point A lies between points B and D. Let E be the midpoint of line segment BC and F the point on side AB such that CF is perpendicular to AE.

Prove that |BF| = 2|AD|.

Let DA = x and DC = y, so that AB = x + y and DB = 2x + y.
Let CF and AE intersect at G and let angle CDB = d.
Let the mid-point of BD be M, so that EM is parallel to CD, EM =y/2 and
angle EMB = d (using the similar triangles BEM and BCD).

AM = DB/2 - x = y/2, which shows that AM = EM and that triangle EAM is isosceles with angle EAM = d/2 (half its exterior angle EMB).

Now, in triangle AFG, angle AFG = 90 – d/2 so that in triangle CFD,
angle FCD = 180 – d – (90 – d/2) = 90 – d/2, which proves triangle FCD to be isosceles. Thus DF = y, giving BF = 2x = 2 AD

Posted by Harry on 2009-11-26 12:34:50