Posted by **m** on Sunday, July 19, 2009 at 4:28am.

: Hi helper, thanks for the help, this website is very good and it would not of been possible without your assistance.

Here is a question i would like you to solve for me this time

.......................................

let ABC be a triangle. Two straight lines which are parallel to the side AC divide the triangle into three figures of equal are. In what parts is the side AB divided by the lines, if AB= 10 cm

- math -
**MathMate**, Sunday, July 19, 2009 at 10:25am
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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