# 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.

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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.

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.