Make a figure with two squares: ABCD and EFGC, with DCG perpendicular to ECB. (The squares are not [necessarily] the same size.) Add lines DE, DF and AE. DF crosses AE at point H. AE crosses DC at point I. DF crosses EC at point J. Show that the areas of DHE and HJCI are equal

I couldn't find a simple geometric solution.  The lengths involved are pretty crazy and there aren't many similar triangles to work with.  So I went analytic.

Put the first square with A=(a,-a), B=(0,-a), C=(0,0), D=(a,0)
the second square E=(0,b), F=(-b,b), G=(-b,0)

Line DE = y=(-b/a)x + b
Line DF = [-b/(a+b)](x-a)
Line AE = [-(a+b)/a]x + b

for simplicity let a²+ab+b² = e

Point H = (ab²/e , a²b/e)
Point I = (ab/(a+b),0)
Point J = (0,ab/(a+b))

Long story short:
the areas both come out to a²b²/(2e)

Posted by Jer on 2014-02-16 20:04:20