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.

Guest Mar 4, 2021

#1**0 **

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 :)

SparklingWater2 Mar 4, 2021