In the accompanying figure, i am told that the area of trapezoid BCGF equals the area of trapezoid FGED. With BC = a, and DE = b, i am to prove that the length of FG is sqrt((a^2 + b^2) / 2). Here is the suggested solution provided by my book:
The areas of the three triangles are similar and therefore proportional with a^2, FG^2, and b^2. Since the area of triangle FGA is the mean of the areas of triangles DEA and BCA, we must have FG^2 = (a^2 + b^2) / 2.
Please help me in understanding why the areas of the triangles are proportional with a^2, FG^2, and b^2, and why FGA is the mean of the two other triangles. Thank you.
if $DE$ is parallel to $FG$ is parallel to $BC$
Then triangle $ABC, ADE, AFG$ are similar triangles.
The area of each then is proportional to the square of the base.
We can say that the area of triangle $ABC = \alpha BC^2$ and the area of triangle $AFG = \alpha FG^2$
Where $\alpha$ is some constant that depends on the shape of the triangle. We don't know it, but we don't need to know it.
Area of Trapezoid $BCGF = $ area triangles $ABC - AFG = \alpha (BC^2 - FG^2)$
Area of Trapezoid $FGED = $ area triangles $AFG-ADE = \alpha (FG^2 - DE^2)$
$BC^2 - FG^2 = FG^2 - DE^2\\ 2FG^2 = BC^2 + DE^2$