In square ABCD, E is the midpoint of BC, and F is the midpoint of CD. Let G be the intersection of AE and BF. Prove that DG = AB.
I've been sitting for 15 minutes and I still can't figure this out :(:(:(. HELP!!!!!! and thank you!
+128475 Let the side of the square be, S
Let A = (0,0) B = (0, S) C = (S, S) D = (S,0) E = (S/2, S) F = (S, S/2)
The slope of the line containing the segment AE is [S -0 ] / [S/2 - 0] = S/[S/2] = 2
So the equation of the line containing this segment is
y - 0 = 2(x - 0)
y = 2x (1)
The slope of the line containing the segment BF is [ S - S/2] / [ 0 - S] = [S/2] / [ -S] = -1/2
So.....the equation of the line containing this segment is
y - S = -(1/2)(x - 0)
y = -(1/2)x + S (2)
Setting (1) = (2) we can find the x coordinate of G
2x = -(1/2)x + S
(5/2)x = S
x = (2/5)S
And the y coordinate of G is y = 2x = 2(2/5)S = (4/5)S
So....by the distance formula, DG is
√[ (S - (2/5)S)^2 + ((4/5)S - 0)^2 ] =
√[ [(3/5)S ]^2 +[ (4/5)S]^2 ] =
√ [ (9/25)S^2 + (16/25)S^2 ] =
√ [ (25/25)S^2 ] =
√[ 1S^2 ] =
S
Which is the same length as AB = the side length of the square
