By: Diana Brown
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.
Click here to open a GSP file and drag point P for yourself and see what happens to each length.
Notice that in each example and no matter where you drag point P that
(AF)(BD)(EC) = (FB)(DC)(EA), and if we divide both sides by (FB)(DC)(EA) then we get:
To prove this lets take Dr. Wilsonís advice and draw parallel lines to produce similar triangles.
You can see that there are several transversals to the segments GH and BC. Because of the parallel lines and transversals we have several congruent angles.† There are also several congruent angles due to the angles being vertical.† These congruent angles are also producing similar triangles, which should produce some ratios to help me prove the statement:
Looking at the above diagram you can see that the similar triangles are:
AHF & BCF, AEG & BCE, AGP & BDP, CDP & AHP
From these similar triangles we
derive the following ratios:
(1) AF/FB = AH/BC
(2) CE/EA = BC/AG
(3) AG/BD = AP/DP
(4) AH/DC = AP/DP
From statement 3 and 4 we can conclude that (5) AG/BD = AH/DC and (6) BD/DC = AG/AH
If we multiply statements (1), (2), and (6) together we get:
Notice that the right side of the
equation cancels to one, and thus we have our original statement:
Show that 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. When is it equal to 4?
Lets look at some examples of different ratios of the areas of triangle ABC & DEF:
Diagram A††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††† †††††††††††††††††† Diagram B
Diagram C††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††† †††††††††††††††††† Diagram D
As you can see in the diagrams that when Point P is moved about in triangle ABC and DEF that the ratio of the areas of triangle ABC and triangle DEF is always greater than or equal to 4.† You can click here to open a GSP sketch of the above diagrams to move point P for yourself.
Notice that Diagram D is when the ratio of the areas of triangle ABC and triangle DEF is equal to 4, why is this.
My first initial guess is that Point P must be at one of the centers of the triangles.† If I draw a line segment from the vertices of triangle ABC to the midpoint of the opposite side the line segment does join to the vertices of triangle DEF, so therefore point P is at the centroid of Triangle ABC.