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.

Let explore (AF)(BD)(EC) and (FB)(DC)(EA) for various triangles and various locations of P.
When P is on the side of the triangle ABC, a line segment connecting the vertex and the opposite side divides
the triangle ABC into two. But if P is inside the triangle ABC six triangles are formed. the line segment
AF, FB, Bd, Dc, CE, and EA intersect with two line segments from the vertex P to form the triangle
APF. APE, FPB, EPC BPD and CPD.
When the vertex  of the triangle ABC is dragged, the inside triangles change shapes.
Similar pairs of triangles are observed.

Conjecture:

(AF)(BD)(EC) = (FB)(DC)(EA)

Prove:
To prove the above conjecture, let examine the similarity of the inside triangles of the construction below.
Let the triangle ABC an isosceles triangle. If P is on the segment AD which bisect the segment BC, then the triangle
ABD and ACD are congruent by SAS theorem.
So is the triangle BEC and BEA.
If two triangle are similar, then the measures of the corresponding altitude are proportional to the measure of the corresponding sides.
Based on the above theorem I can conclude the triangle BPD is similar ot the triangle
CPD, the triangle BPF is similar ot the triangle CPF, and the triangle APF is similar ot the
triangle APE,

Another aspect to prove the similarity between the pairs of triangles is to examine their
areas.

The ratio: (AF)(BD)(EC) / (FB)(DC)(EA) =1

When P is inside triangle ABC, the ratio of the areas of triangle ABC and triangle DEF is
always greater than or equal to 4.
the reason why the ratio is always greater than 4 is due to the fact that the largest area of
the triangle DEF is one fourth of the triangle ABC.
The triangle DEF reaches the largest possible area when DEF is an isosceles triangle
and the remaining triangles are also isosceles triangles.

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