Zach.J.Dragone wrote:An equilateral triangle ABC is inscribed in square ADEF, forming three right triangles: ADB, ACF and BEC. What is the ratio of the area of triangle BEC to that of triangle ADB?

A. 4/3
B. √(3)
C. 2
D. 5/2
E. √(5)

Spoiler: C

Draw the figure:

Since AD=AF and AB=AC, ∆ADB are ∆ACF are CONGRUENT.

Let BE=2 and DB=x.
The result is the following figure:

Since BE=CE=2, ∆BEC is a 45-45-90 triangle.
In a 45-45-90 triangle, the sides are in the following ratio:
s : s : s√2.
Thus, BC = 2√2.

∆BEC:
Area = (1/2)(2)(2) = 2.
.
∆ABC:
The area of an equilateral triangle = (s²/4)(√3).
Thus, the area of ∆ABC = (2√2)²/4 * √3 = 2√3.

Square ADEF:
The area can be represented TWO WAYS:
Area = s² = (x+2)² = x² + 4x + 4.
Area = ∆ADB + ∆ACF + ∆ABC + ∆BEC = (1/2)(x+2)(x) + (1/2)(x+2)(x) + 2√3 + 2 = x² + 2x + 2√3 + 2.

Since the area must be the same in each case, we get:
x² + 4x + 4 = x² + 2x + 2√3 + 2
2x = 2√3 - 2
x = √3 - 1.

∆ADB:
Area = (1/2)(x+2)(x) = (1/2)(√3 - 1 + 2)(√3 - 1) = (1/2)(√3 + 1)(√3 - 1) = (1/2)(3 - 1) = 1.

Resulting ratio:
∆BEC/∆ADB = 2/1 = 2.

The correct answer is C.

Cool problem, but too complex for the GMAT.

Thanks for the response Mitch

Actually, this problem isn't so bad. For some reason I just wasn't seeing that we should be drawing the squares diagonal through the equilateral and isosceles triangle. Once I recognized that solving was pretty easy.