The figure below shows a trapezium ABCD, where AB : DC = 2 : 3 is the mid-point of AB. Straight lines DE and DB meet AC at F and G respectively. Find the radio of AF : FG : GC.
Probably a way better "geometric" way to do this
Let
A = (3,4)
B = (7,4)
C = (6,0)
D = (0,0)
E = (5,4)
The equation of the line containing DE is y = (4/5)x (1)
The equation of the line contining DB is y = (4/7)x (2)
The slope of the line containing AC is [ 4 - 0] / [ 3 - 6 ] = -4/3
And the equation of this line is y = (-4/3)(x - 6) = (-4/3)x + 8 (3)
The intersection of (1) and (3) will give the x coordnate of F
So
(4/5)x = (-4/3)x + 8
(4/5)x + (4/3)x = 8
([ 12 + 20] / 15)x = 8
([ 32] / 15 )x = 8
x = 15 * 8 / 32
x = 15/4 = 3.75
And y = (4/5) (15/4) = 3
So F = ( 3.75, 3)
And the intersection of (2) and (3) will give the x coordinate of G
So
(4/7)x = (-4/3)x + 8
(4/7)x + (4/3)x = 8
([ 12 + 28] / 12) x = 8
( [ 40 ]/ 21) x = 8
x = 8 * 21 / 40
x = 21/ 5 = 4.2
And y = (4/7)(21/5) = (4/5)* 3 = 12/5 = 2.4
So G = ( 4.2, 2.4)
And AF = √[(3 - 3.75)^2 + ( 4- 3)^2 ] = √[.75^2 + 1] = 1.25
And FG = √[ (3.75- 4.2)^2 + ( 3 - 2.4)^2 ]= √[[.45^2 + .6^2] = √.5625 = .75
And GC = √[(4.2 - 6)^2 + 2.4^2 ] = √ [ 1.8^2 + 2.4^2 ] = √9 = 3
So
AF : FG : GC =
1.25 : .75 : 3 =
5 (.25) : 3 (.25) : 12 (.25) =
5 : 3 : 12