ABCD is a parallelogram. E is a point on AB such that AE:EB = 1:2, F is the mid-point of CD. AF meets DE at G and BF meets CE at H. When expanded on both sides, GH meets AD at M and BC at N. If the area of ABCD is 400, find the sum of the areas of the triangles ∆CNH and ∆DGM.
+592 This one is very complicated. The way I did it was graphing, and wrote D as the origin point, so everything is positive to make it simpler. we have to find the slope and intercept of DE and AF to get the intersection:G. we can do the same for BF and EC to get intersection: H. then, we can find the slope and intercept of GH to get M and N. the height of triangle CNH=20-x coordinate of h, and the base will be y coordinate of N. similarly, the height of triangle DGM=x coordinate of G and the base is 20-y coordinate of M, so we can mulitply them out:
equation for DE:$\frac{20-0}{\frac{20}3-0}=3$
y=3x+0
equation for EF:$\frac{20-0}{0-10}=-2$
y=-2x+20
G:0=5x-20
5x=20
x=4
y=12
G=(4,12)
equation for BF:$\frac{20-0}{20-10}=2$
y=2x-20
equation for EC:$\frac{20-0}{\frac{20}3-20}=-\frac32$
y=$-\frac32$x+30
H:0=$\frac72$x-50
$\frac72$x=50
x=$\frac{100}7$
y=$\frac{60}7$
H=($\frac{100}7,\frac{60}7$)
you can do the rest by yourself :)
