Since 3(AD) = DB, then D must be (1/4) the distance from A to B
We can figure its coordinates as
[ 0 + (1/4)(9-0) , 0 + (1/4)(6-0) ] = ( 9/4, 6/4) = (9/4, 3/2)
Likewise, we can find E as
[ 9 + (1/4)(6-9) , 6 + (1/4)(12 - 6) ] = [ 9 - 3/4 , 6 + 6/4 ] = [ 33/4 , 6 + 3/2 ] = [33/4 , 15/2 ]
Finally, we note that F is 3/4 of the distance from A to C
To find its coordinates we have
[ 0 + (3/4)(6-0) , 0 + (3/4) (12-0) ] = [ 18/4, 36/4 ] = [ 9/2, 9]
To find the area of ABC we can use this formula where
Let (0,0) = (x1, y1) (9,6) = (x2, y2) (6,12) = (x3, y3)
Area = (1/2) [ x1 (y2 - y3) + x2 (y3 - y1) + x3(y1 - y2) ]
(1/2) [ 0 ( 6 - 12) + 9(12 -0) + 6(0 - 6) ] = (1/2 [ 9 *12 + 6(-6) ] =
(1/2) [ 108 - 36 ] =
(1/2) 72 =
36
Likewise
Let (9/4, 3/2) = (x1,y1) (33/4, 15/2 ) = (x2, y2) (9/2, 9) = (x3, y3)
Area = (1/2) [ (9/4) ( (15/2) - 9) + (33/4) (9 - 3/2) + (9/2)(3/2 - 15/2) ] = 63/4
Area of DEF to Area of ABC =
(63/4) / 36 =
63 / 144 =
7 / 16