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Nagel Point of the Medial Triangle: What Is It About?
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Copyright © 1996-2008 Alexander Bogomolny
The applet may suggest the following problem (Problem 188, proposed by Michael Wolterman, Math Horizons, April 2005, p. 33, MAA):
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Let EDF be the medial triangle of |
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Two simple facts important to solving the problem stand out:
- Two tangents from a point outside a circle to the circle are equal.
- The three triangles AEF, BDF, and CDE are just translations of each other.
Thus, for example, the two tangents from E to the incircle of 
CDE are equal as are the tangents from F to the incircle of BDF. In fact all four are equal. In particular,
| (1) | ER = FQ. |
Similarly,
| (2) | DR = FP. |
| (3) | DQ = EP. |
Ceva's theorem tells us that the lines DP, EQ, FR are indeed concurrent.
The unexpected fact about this configuration is that the point of concurrency M is none other than the Nagel point of the medial triangle DEF. In other words, the points P, Q, R are the points where the excircles of DEF touch its sides! This follows from the fact that the points P, Q, R are perimeter-splitters (as is the case with the Nagel point.) I.e., for example,
| EF + ER = DF + DR. |
But this is the characteristic property of the points of tangency of the excircles and, therefore, of the Nagel point. And the proof is complete.
The applet's diagram can be used to illustrate the fact of isotomic conjugacy of the Nagel and Gergonne points.
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Let P', Q', R' be the points of tangency of the incircle of DEF on the sides EF, DF, and ED. Then since triangles EDF and EDC are symmetric in the midpoint of ED, the same holds for the points of tangency R and R' of their incircles with ED. However, as we've seen, R is also the point of tangency of the excircle of DEF opposite vertex F. Treating pairs the Q/Q' and P/P' similarly, we see that indeed the Nagel and Gergonne points of DEF are isotomic conjugate.
The consequence of the above is that the Nagel point of a triangle coincides with the incenter of its anticomplementary triangle. In the current configuration then M coincides with the incenter I of ABC.
Copyright © 1996-2008 Alexander Bogomolny
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