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Larousse Charlot

Proof Using Geometer’s Sketchpad


Using Geometer’s Sketchpad, we are going to explore the relationship between two triangles.  However, we going through the exploration from a proof approach so that we justify our statements and conclusion.


To have our starting triangle, let us suppose that we have three distinct noncollinear points A, B, and C such that we obtain the triangle ∆ ABC.


Noncollinear points




Now, let the points D, E, F be midpoints of the segments AB, BC, and AC respectively.

So, the segments AD Ň DB, BE Ň EC, AF Ň FC.

From there we know that D, E, F are between the points AB, BC, and AC, by axioms of betweenness.



If connected the three medians and make a triangle, we have what is called a median triangle.  First, let us have the line segment of DE.  DE is the mid segment of the triangle ∆ ABC. 

By definition of mid segment of triangle, we have

DE = ½ AC


EF = ½ AB

DF = ½ BC

From our observation here, we see that the triangles ∆ ABC and ∆ DEF have a ration of 2:1.  Hence, the perimeters of the triangles would have the same ration.

When the length of each were measured, we obtained the following

However, the ratio of the area of the triangles is not 2:1.  Why?  Well, because the measurements will be squared.  Hence to get the right ratio, the ratio, too, has to be squared.  So, the ratio for the area of the triangles is 4:1


So, we can state that the triangles are congruent proportionately.  Therefore, if the original triangle is an equilateral, the median triangle too will be an equilateral triangle.  The same applies for isosceles, obtuse, and right triangle[1].





[1] A triangle is a right triangle if one of its angles is congruent to its supplementary angle.  In Euclid’s geometry, such angle measures 90ľ.  The median of a triangle is a ratio of its original triangle.