This assignmnet involves beginning with any arbitrary triangle ABC. We select a point P inside the triangle and construct lines AP, BP, and CP extended to their intersections with the opposite sides in points D, E, and F respectively. This is shown in the diagram below.
In this particular triangle, (AF)(BD)(EC) = (FB)(DC)(EA). This is true for various other triangles and various other locations of P. A conjecture can be made that for all triangles ABC with P, D, E, and F as described above, (AF)(BD)(EC) = (FB)(DC)(EA).
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a proof of this conjecture.
The above conjecture is true even when point P is outside the triangle.
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When P is located inside triangle ABC, the ratio of the areas of triangle ABC and DEF is always greater than or equal to 4.
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