SOLUTION: Show that the medians of any triangle divide the triangle into six smaller triangles of equal area. Are any of the small triangles necessarily congruent to each other?

Question 1155225: Show that the medians of any triangle divide the triangle into six smaller triangles of
equal area. Are any of the small triangles necessarily congruent to each other?
Answer by Edwin McCravy(20055) (Show Source): You can put this solution on YOUR website!

 

First we use these two theorems:
1. the medians are concurrent and intersect two thirds the distance from the
vertex to the midpoint of the opposite side. 

2. The area of a triangle is one-half the product of any two sides multiplied by
the sine of their included angle.

∠AMD = ∠CME because they are vertical angles
AM = 2ME because AM is 2/3 of AE
MC = 2MD => MD = 1/2∙MC because MC is 2/3 of CD

Area of ΔAMD = (1/2)ME∙MC∙sin(∠CME)

Area of ΔCME = (1/2)AM∙MD∙sin(∠AMD) = (1/2)(2ME)(1/2∙MC)∙sin(∠CME) =
(1/2)ME∙MC∙sin(∠CME) by substitution.
So ΔAMD and ΔCME have the same area.

Similarly ΔBMD and ΔCMF have the same area, and so do ΔAMF and ΔBME,
as well as ΔBMD and ΔCME.

Next we use the theorem:

If two triangles have the same base and altitude lengths, they have the same area.

That follows immediately from the formula A = (1/2)∙base∙altitude

We draw in the perpendiculars from the median to the three sides of ΔABC, each
of which is a common altitude of two of the 6 triangles.

 

Since the medians are bisectors of the sides they are drawn to, 
ΔAFM and ΔFCM have the same measure of base and altitude (in green),
ΔADM and ΔBDM have the same measure of base and altitude (in green). 
ΔBEM and ΔCEM have the same measure of base and altitude (in green).

The information above is sufficient to show that all 6 triangles have the same
area.  None are necessarily congruent.

Edwin

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