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# In-depth solutions are greatly appreciated!

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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!

AnonymousConfusedGuy  Mar 25, 2018
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#1
+85759
+2

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

CPhill  Mar 25, 2018
#2
+925
+2

Thank you so much!

AnonymousConfusedGuy  Mar 25, 2018

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