Similar Triangles




Brock F. Miller



This final investigation is to see the conjectures and proofs that can be made from looking at an arbitrary point, P, in the interior of D ABC and the segments connecting the vertices, P, and the opposite sides.




Consider the products: (AF)(BD)(CE) and (FB)(DC)(EA) for various points P in the interior of D ABC.

CASE I: P is the centroid of D ABC.

Conjecture: If P is the centroid of D ABC then,




By definition of centroid we can conclude that AF = BF, AE = EC, and BD = CD.

By direct substitution we can conclude:


1 = 1

Therefore we have proved that if P is the centroid then the ratio of the indicated sides is equal to 1. Now this is not exciting but is a good place to start. Now let's prove our conjecture for any D ABC with P in the interior. To prove this we need to use similar triangle geometry. To aid in finding the necessary similar triangles we will run two lines parallel to segment CF and with one intersecting point A and one intersecting point B. We will then extend segments AD and BE to intersect these lines and label the points J and K.


There are numerous pairs of similar triangles in this figure we will focus on four sets of them to prove our conjecture.

By using vertical angles and pairs of angles being congruent by alternate interior angles we can conclude two similarity statements and respective ratio of corresponding sides.







By corresponding angles we know that <AJB is congruent to <FPB. By the reflexive property we can conclude that <PBF is congruent to <JBA. Therefore by AA similarity we can conclude:



with corresponding proportion

which implies that

Likewise by a pair of corresponding angles and a use of the reflexive property we can conclude another similarity by AA.






with corresponding ratios of

which implies,


Before we go any further let's review the four proportions we are going to use to prove our conjecture:






From parts I and II and the multiplicative property of equality we obtain

With simplification of the left-hand side of the equal sign we get the equivalent equation of:

From Parts III and IV we can substitute for BK and AJ and obtain the equivalent equation.


Simplifying the complex fraction creates the equation



Using the Commutative Property of equality we can manipulate the equation to

Multiplying both sides of the equation by the inverse of the right hand side

Which simplifies to

Which is equivalent to the conjecture we wanted to prove:

Therefore when P is in the interior of D ABC, it can be concluded that the ratio of the indicated sides is equal to a value of 1.


We have proved the case when P is in the interior of the triangle but what if P is in the exterior?

I will leave this to you to make your conjecture and proof. GSP 4.0 users for a picture and press the "ANIMATE" button to see P at various positions in the exterior of the triangle.

CLICK HERE for Picture


We have looked at two cases of P on the interior of the triangle and left you to think about the possibilities of P on the exterior. Let's consider the areas of D ABC and D DEF when P is in the interior.

What was asked to show was that the ratio of the areas of D ABC and D DEF is greater than or equal to four and we were asked to find when the areas were equal to four. Here is a concrete example of this conjecture.

Clearly if we divide 25.58 by 6.14 we get a value that is greater than four. GSP 4.0 Users use the picture to move P around the interior to see other areas. To see picture - CLICK HERE


Notice that the only area is that is subject to change when P is in the interior is that of triangle DEF. This should make sense that the movement of P in the interior will only change the magnitude of the interior triangle DEF.


CONJECTURE: The ratio of areas of the two triangle will equal four when P is the centroid of the triangle.

Let P be the centroid of triangle ABC. Let D, E, F be the midpoints of the sides of triangle ABC



By the Midpoint Connector Theorem we know :

By SSS similarity we can conclude that D ABC ~ D DEF.

Therefore the heights to the corresponding sides must be proportional by 1/2.




Now by rule for area of a triangle the area of each triangle followed by the ratio of areas would be:

Which will simplify to:

By substitution for EF and h2 we obtain:

Which simplifies to prove the ratio equals 4.

I hope this look into the situation of creating similar triangles within a given triangle has been as enjoyable for you to read as it was for me to investigate. I would enjoy to see other proofs for this topic and encourage you to look at different angles then the ones I have taken. Please email me any questions or comments from my web page.


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