A. Consider any triangle ABC. Select a point P inside the triangle and draw lines AP, BP, and CP extended to their intersections with the opposite sides in points D, E, and F respectively. Explore (AF)(BD)(EC) and (FB)(DC)(EA) for various triangles and various locations of P.
From the above four triangles, we can see that (AF)(BD)(EC) = (FB)(DC)(EA).
B. Prove the above.
We can get the following figure by constructing lines through B and C parallel to AD and extending their intersections through CF and BE. The three segments BX, AD, and CY are parallel.
Because of equal vertical angles and equal interior angles, triangle AFP ~ BFX, and triangle CEY ~ AEP. From this similarity, we can get the following ratios:
AF/BF = AP/BX and CE/AE = CY/AP
Because of the common angles and equal corresponding angles, triangle BDP ~ BCY and triangle BCX ~ DCP . And the following ratios:
BD/BC = DP/CY and BC/DC = BX/DP.
Now let's multiply the left and the right sides of these equations to get
(AF/BF)(BD/BC)(BC/DC)(CE/AE)=(AP/BX)(DP/CY)(BX/DP)(CY/AP) and simplifying we get,
(AF/BF)(BD/DC)(CE/AE) = 1 and this means that (AF)(BD)(CE) = (BF)(DC)(AE) as required.
And this ratio will also hold true even when the point P is outside of triangle ABC as shown below.
The above two figures show that the ratios are greater than or equal to 4.
The ratio is exactly 4 when P is the centroid of the triangle as shown below.
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