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\(ABCD\) is a square with side length 1\(E\) and \(F\) are the mid-points of \(AD\) and \(AB\) respectively. \(G\) is the  intersection point of \(CF\) and \(BE\). Find the length of \(DG\).

 

 Nov 16, 2020
 #1
avatar+114221 
+2

Let

D = (0, 0)

C = (1,0)

E = (0, 1/2)

F = (1/2, 1)

B = (1,1)

 

Slope of line connecting EB = [ 1 -1/2 / [ 1 - 0 ] =   1/2

Equation of lime  connecting EB    ....y = (1/2) x + 1/2      [ 1 ] 

 

Slope of line connecting CF   =  [ 1 - 0 ] / [1/2 -1 ] = 1 / -1/2   =  -2 

Equation of line  connecting CF  ......y = -2 ( x - 1) ..... y = -2x + 2    [2]

 

x intersection [1] and [2]  = x value of G

 

(1/2)x + 1/2  = -2x + 2

(5/2)x = 3/2

5x = 3

x = 3/5

 

And using either line to find the y value of G

 

y= -2(3/5) + 2 =  -6/5 + 10/5 =  4/5

 

G = ( 3/5 , 4/5)

 

 

DG =  sqrt  [ (3/5)^2  + (4/5)^2 ]  = sqrt  [9/25 + 16/25 '  =  sqrt  [25/25]  = sqrt [ 1 ]  =  1

 

 

cool cool cool

 Nov 16, 2020
 #2
avatar
+1

BE and CF intersect at 90º

 

∠DCG = ∠CBG = arctan(2 / 1)

 

CG = sin(∠CBG * BC)

 

Use the law of cosines to calculate the length of line DG.

 

 Nov 16, 2020
 #3
avatar+846 
0

laughlaugh I'm laughing at my own error.

 

Instead of   ∠DCG = ∠CBG = arctan(2 / 1)    should be  ∠DCG = ∠CBG = arctan(1 / 0.5)

 Nov 16, 2020

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