4.2.4 Introduction to Coordinate Proof (continued)
In Method 2, the triangle is again positioned on a coordinate plane such that B is at the origin, A is on the y -axis and C is on the x -axis. But in this proof, the lengths of the sides are not m and n . Instead, let AB = 2 j and let BC = 2 n . The point of identifying these lengths as “twice the variable” is so that the coordinates of the midpoints will not contain fractions. By using the midpoint formula, the base of △ DBE is found to be n and the height of △ DBE is found to be j . Now find the area of each triangle and compare the areas to prove the conjecture. Now find the area of the two triangles. Since △ ABC has base m and height n , the area of △ ABC is mn /2 units 2 . Since △ DBE has base m /2 and height n /2, the area of △ DBE is mn /8 units 2 . Therefore, since mn /8 = (1/4)( mn /2), the area of △ DBE is one-fourth the area of △ ABC .
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