+1450 In square ABCD, E is the midpoint of line BC, and F is the midpoint of line CD. Let G be the intersection of line AE and line BF. Prove that DG = AB.
Thanks so much!
+128407 Since we can use any scale factor for the square that we desire, we can derive an algebraic solution
Let A = (0, 0)
B = (0, a)
C = (a, a)
D = (a, 0)
E = (a/2, a)
F = (a, a/2)
AB = a
The slope of AE is (a) / ( a/2 - 0 ) = a / (a/2) = 2
The slope of BF is ( a - a/2) / ( 0 - a) = -(1/2)a / a = -1/2
The equation of the line containing segment AE is y = 2x
The equation of the line containing segment FB y = (-1/2)x + a
Set these equal to find the x coordinate of G
2x = (-1/2)x + a
(5/2)x = a
x = (2/5)a
And the y coordinate is y = 2[(2/5)a] = 4/5a
So....G = (2/5 a, 4/5 a)
And DG = √ [ (2/5 a - a)^2 + (4/5 a)^2 ] = √ [ ( 3/5 a)^2 + (4/5 a)^2 ] =
a√[ 3^2 + 4^2 ] / 5 = a √25 / 5 = (5 / 5 ) a = a = AB
