In triangle ABC, MN || BC and MN and MN bisects the area of triangle ABC. If AD = 10, find ED, if AE is the height of triangle AMN.

Since triangles ABC and AMN are similar and area of AMN is
half of the area of ABC, we'll have;
Area of AMN/Area of ABC = 1/2
(1)
Let ED = x => AE = AD - ED = 10-x
Area of AMN/Area of ABC
= (10-x)^2/10^2 (2)
We'll equate (1) and (2) and we'll
get:
(10-x)^2/10^2 = 1/2
We'll cross multiply:
2(10-x)^2 =
100
We'll expand the binomial and we'll multiply by 2:
200 - 40x +
2x^2 - 100 = 0
x^2 - 20x + 50 = 0
We'll apply quadratic
formula:
x1 = [20+sqrt(400 - 200)]/2
x1 = 10+5sqrt2
x2 =
10-5sqrt2
Since x has to be smaller than 10, we'll keep only the second value,
namely x = ED = 10-5sqrt2.