We have a square ABCD with sides = 1, and BE=1, and AE=CE.
What is the area of triangle ABE?
How can I solve this?
I found two ways to solve this and then stop. this is the shorter of the two methods. See the modified diagram below.
from the description we see that $\displaystyle \triangle AEC$ is isosceles. thus if we extend the line BE, it will cut the diagonal at a right angle. from this we can the angles that you see filled in.
now we simply use the formula $\displaystyle \mbox {Area} = \frac {1}{2}(BE)(AE) \sin 135^{ \circ} = \frac { \sqrt {2}}{4}$
if you have any questions say so. i was kind of rushing through this and didn't explain much