Posted by m on Sunday, July 19, 2009 at 4:28am.
Let the parallel lines DD' and EE' cut BA at D,E and BC at D', E'. Thus the sides BA and BC are now named BDEA and BD'E'C.
The question requires that the figures BDD', DD'E'E, and EE'CA have the same area. Thus the area of the triangles BDD', BEE' and BAC are in the ratio of 1, 2 and 3.
The area of similar triangles are proportional to the square of the sides, so
(BD/BA)2:(BE/BA)2:(BA/BA)2
=(1/3):(2/3):(3/3)
Therefore
BD=BA sqrt(1/3)
BE=BA sqrt(2/3)
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