Nagel Point of the Medial Triangle
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A Mathematical Droodle
Explanation
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Copyright © 1996-2012 Alexander Bogomolny
The applet may suggest the following problem (Problem 188, proposed by Michael Wolterman, Math Horizons, April 2005, p. 33, MAA):
Let EDF be the medial triangle of ΔABC. The incircles of triangles AEF, BDF, and CDE are tangent to EF, FD, and DE at P, Q, R, respectively. Prove that DP, EQ, FR are concurrent. Furthermore, the point of concurrency is the Nagel point of DEF.
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 ΔCDE are equal as are the tangents from F to the incircle of ΔBDF. In fact all four are equal. In particular,
Similarly,
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 ΔDEF 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.
Let P', Q', R' be the points of tangency of the incircle of ΔDEF 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 ΔDEF opposite vertex F. Treating pairs the Q/Q' and P/P' similarly, we see that indeed the Nagel and Gergonne points of ΔDEF are isotomic conjugate.
