Isosceles on the Sides of a Triangle: What is this about?
A Mathematical Droodle
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Copyright © 19962018 Alexander Bogomolny
The applet's purpose is to suggest the following statement:

On the sides of ΔABC erect isosceles triangles AFB, BDC, and CEA. Prove that the perpendiculars through A, B, C to EF, FD, DE, respectively are concurrent.

Proof
The proof is straightforward. Consider three circles cD, cE, cF centered at D, E, F with radii DB = DC, EC = EA, and FA = FB, respectively. cD and cE meet in C and some other point, say, c so that Cc is their common chord and serves therefore their radical axis. The chord Cc is orthogonal to the line of centers, DE. That is, Cc coincides with the perpendicular from C to DE. Similarly, the other two perpendiculars, serve as the radical axes of the other two pairs of the circles. But the radical axes of three circles concur in a point, the radical center of the circles. So do the perpendiculars at hand.
References
 R. Honsberger, Mathematical Delights, MAA, 2004, pp. 1920
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Activities
Contact
Front page
Contents
Geometry
Copyright © 19962018 Alexander Bogomolny