中二數學的問題(10分)

Let FC = y.
Therefore, FG = y (G is the point originally at C).
BF = AD - FC = 12 - y.
BG = DC = 10.
For triangle BGF, by Pythagoras thm.,
BG^2 + FG^2 = BF^2
10^2 + y^2 = (12-y)^2
100 + y^2 = 144 - 24y + y^2
24y = 44, so y = FC = 11/6.
Let M be the point on AD such that FM is perpendicular to AD.
Therefore, FM = DC = 10.
BE = BF = 12 - y = ED = EM + MD
MD = FC = y
so 12 - y = EM + y
EM = 12 - 2y.
For triangle EFM, by Pythagoras thm. again,
EF^2 = FM^2 + EM^2
EF^2 = 10^ 2 + (12 - 2y)^2
= 100 + (12 - 11/3)^2 = 100 + (25/3)^2 = 100 + 625/9 = 1525/9
therfore, EF = sqrt(1525/9) = (5sqrt61)/3.