Using the measurements of the lengths of the segments, we find that
Connect to GSP to convince yourself. ABCsegment.gsp
In order to prove this, construct two lines through B and C that are parallel
to AD.
Since triangle PBD is similar to triangle QBC and triangle PCD is similar
to triangle RCB
(corresponding angles are congruent), we get
Since triangle RFB is similar to triangle PFA and triangle QEC is similar
to triangle PEA (alternate interior angles and vertical angles are congruent),
we have
From the similar triangles we can get the following ratios:
Manipulating the first two ratios we get
Thus, using the previous ratios with this new one, we have
Simplifying the right side of the equation, we get
From the following picture we can see that even when the point P is outside
the triangle ABC, the ratio still holds.
Visit this GSP link to manipulate the figure
for yourself.