Consider any triangle ABC. Let P be any point inside the triangle and draw lines AP, BP, and CP. Let the points of intersection with the opposite sides be D, E, and F respectively.

Using the measurements of the lengths of the segments, we find that

(AF)(BD)(CE) = (FB)(DC)(EA) or

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.

Another interesting point is that when P is inside triangle ABC, the ratio of the areas of triangles ABC and DEF is always greater than or equal to 4.

For a demonstration connect to GSP .

One might ask what causes the ratio to be exactly 4? When P is constructed as the centroid of triangle ABC and triangle DEF is constructed using the midpoints of the sides of ABC, the ratio equals 4.

See the figure in GSP.

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