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?

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- Jun 27th 2007, 05:39 PMthumpin_termisFinding an area of a triangle:
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? - Jun 27th 2007, 06:05 PMJhevon
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