If we join diagonal BD of parallelogram ABCD , we get another triangle ABD, such that
Area of || gram ABCD = 2 Area of tri (ABD) ( since both have same base and lie bet same parallel lines) ---- 1
Area of || gram ABCD = 3 Area of tri (ABE) (Given in the question)---- 2

From 1 & 2,
2 Area of tri (ABD) = 3 Area of tri (ABE)
ie, Area of tri (ABE) / Area of tri (ABD) = 2 / 3

tri (ABE) and tri (ABD) are Similar triangles (AA Test of similarity)

Hence , Area of tri (ABE) / Area of tri (ABD) = (AE)^2 / (AD)^2 = 2/3
Therefore, [spoiler]AE / AD = sqaure root of ( 2 / 3)[/spoiler]
whatz the answer?

gmattesttaker2 wrote: In parallelogram ABCD, point E lies on side AD. If the area of the parallelogram
is three times the area of triangle ABE, what is the ratio of the length of AE to
the length of AD?

ABCD can be ANY type of parallelogram.

ABCD:
Let ABCD be a square where s=3.
Area of square ABCD = s² = 3² = 9.

∆ABE:
Since the area of square ABCD is 3 times the area of ∆ABE, the area of ∆ABE = 3.
Thus:
(1/2)(AE)(AB) = 3
(1/2)(AE)(3) = 3
AE = 2.

Resulting ratio:
AE : AD = 2:3.