# Translated Triangles

In ΔABC, D, E, F are the midpoints of sides BC, AC, and AB, respectively. Triangle S is formed by the incenters of triangles AEF, FDB, and ECD; triangle T is formed by the circumcenters of these triangles. Prove that ΔS = ΔT.

The applet below illustrates this problem.

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Solution

In ΔABC, D, E, F are the midpoints of sides BC, AC, and AB, respectively. Triangle S is formed by the incenters of triangles AEF, FDB, and ECD; triangle T is formed by the circumcenters of these triangles. Prove that ΔS = ΔT.

The problem has a very simple solution that shows there are other triangles equal to S and T, the triangles formed by the centroids or the orthocenters of triangles AEF, FDB, and ECD are among the bunch.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

The solution is based on the observation that, say, triangle FDB is obtained from AEF by the translation to the vector AF. This translation moves A to F, E to D, and F to B. It moves other points of triangle AEF into points of triangle FDB. The points in the two triangles that relate to each other in this way are said to correspond under the translation. Thus if X (in ΔAEF) and Y (in FDB) correspond to each other than XY = AF, with the same direction. But obviously, the incenters, circumcenters, orthocenters, and centroids correspond to each other, so that the line segments joining the corresponding centers are all equal and parallel to the side AB. Since the same holds for the centers in pairs FDB/ECD and ECD/AEF, the triangles in question (of the incenters and the circumcenters) are not only equal, the have parallel sides and the same orientation as any of the triangles AEF, FDB, or ECD.

### References

1. I. M. Yaglom, Geometric Transformations I, MAA, 1962, Problem 4