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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!

 

 Jul 4, 2019
 #1
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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

 

cool cool cool

 Jul 4, 2019
 #2
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+1

Thank you so much, CPhill!!!!!

 Jul 5, 2019

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