Crazy Geometry (2)

With ref. to the diagram below:

圖片參考：http://i1191.photobucket.com/albums/z467/robert1973/May11/Crazygeom2.jpg


Let DC = x, AB = (k + 1)x where k > 0. Then AE = x and EB = kx

Also let the height of the trapezium be h.

Now we have AECD being a parallelogram and so △FAE and △FCD are congurent, i.e. they have the same area.

Area of ABCD = [x + (k + 1)x]h/2 = (k + 2)hx/2

Height of △FAE = h/2 since it is congurent with △FCD

Area △FAE = (1/2)x(h/2) = hx/4

Thus area of △FCD = hx/4

With ∠GDC = ∠GBA and ∠GCD = ∠GAB, △GDC ~ △GBA with side ratio = 1/(k + 1) and hence area raitio = 1/(k + 1)2.

Thus GA:GC = k + 1 : 1 and then GA:CA = k + 1 : k + 2

Area of △ABC = (k + 1)hx/2

Since △GBA and △ABC have the same height, their area ratio = GA:CA and so the area of △AGB = [(k + 1)hx/2] [(k + 1)/(k + 2)] = (k + 1)2hx/[2(k + 2)]

So area of △GDC = {(k + 1)2hx/[2(k + 2)]} [1/(k + 1)2] = hx/[2(k + 2)]

Area of △DFG = Area of △FCD - area of △GDC

= hx/4 - hx/[2(k + 2)]

= khx/[4(k + 2)]

Area ratio of △DFG : ABCD is given by:

r = {khx/[4(k + 2)]}/[(k + 2)hx/2]

= k/[2(k + 2)2]

Finally:

dr/dk = (8 - 2k2)/[4(k + 2)4]

To maximize r, dr/dk = 0 and so 8 - 2k2 = 0, giving k = 2 since k > 0

So when AB/CD = 3, the area ratio of △DFG : ABCD is maximum.