## Centroids

On the sides of triangle ABC construct triangles similar to a fixed triangle DEF, as shown. To show that the resulting triangle GHI has the same centroid with the triangle ABC.

The triangle CGA can be considered to result from BCH by rotating the second triangle by the angle at C and taking a homothetical to the resulting triangle. As a result of this consideration corresponding sides of the two triangles are at angle(C). Analogous remark is valid also for the other couples of triangles. One sees then the following consequence.
Draw parallelogram CHBJ. Then triangles CGJ and IBJ are similar to triangle ABC. In fact, angle(JCG) is equal to angle(BCA) and the ratios GC/CJ = AC/CB. By comparing lengths and angles, follows that AGJI is a parallelogram. This in turn implies that BJI is similar to ABC.
The statement on the top follows then by looking at GHI as a result of vector additions to the vertices of ABC. In fact GHI results by adding to corresponding vertices of ABC the vectors CG (to C), GI (to A) and JC (to B). Since these vectors are sides of the triangle CGJ their sum is zero and one can apply the result refered in Centroids_0.html to get the desired result.

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