Question 633638: A point labelled E is drawn on the interior of a square and lines drawn to it from each of the corners. Give an argument that would convince someone else who has studied mathematics that the combined area of the 2 darker shaded triangular regions is half the area of the surrounding square...it may not be { in which case you would have to provide an argument that justi fies this conclusion.
(darker shades are covered by the stars*)
blank space is covered with dots
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*.............****E.....*
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Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
Construct a square of side measure  in the first quadrant of the  plane such that one vertex of the square is at the origin and the two sides adjacent to that vertex are cooincident with the  and  axes.
By definition of a square, the four vertices of the square will be ) , ) , ) , and )
Let point E be represented by the ordered pair )
The altitude of triangle ABE to side AB is simply the coordinate of point E, namely . The measure of the base of triangle ABE is the distance from ) to ) , which is simply . The area of a triangle is given by one-half the altitude times the base, or in the case of triangle ABE, 
Likewise it can be shown that the area of triangle CDE is given by }{2}\ =\ \frac{a^2\ -\ ay}{2})
The sum of these two areas is then:
But the area of the square is  , so the sum of the areas of the two shaded triangles is in fact one-half of the area of the square.
John
My calculator said it, I believe it, that settles it
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