Prove:

First, we create a line through A that is parallel to BC.

This allows us to look at several similar triangles and manipulate the properties of those similar triangles.

The first set of similar triangles is: triangle AHF and Triangle BCF.

Corresponding sides of similar triangles have proportional sides. This allows us to look at the extended proportion:

The second set of similar triangles is: triangle AEG and triangle CEB.

Here we look at the extended proportion:

Triangles AGP and DBP are also similar.

The proportion of the corresponding sides is:

And the last set of similar triangles that we need to look at is

triangle CDP and triangleHAP.

Here the proportion relating the sides is:

Using the last two proportions we see that:

and

Using the Transitive Property, we see that

Using the cross product property of proportions we can rewrite the proportion as:

So we have said:

Multiplying we see:

Which simplifies to:

This holds even when P is outside of the triangle, except when P is on one of the lines that include a vertex and are parallel to the opposite side of the triangle.

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