A Property of the Line IO:
A Bundled Statement and Proof
As usual, let I and O be the incenter and circumcenter, respectively, of triangle ABC. Suppose angle C is 30°, and that the side AB is laid off along each of the other two sides to give points D and E so that
|EA = AB = BD.|
Prove that the segment DE is both equal and perpendicular to IO.
Now, the problem is a mixture of two:
- DE is perpendicular to IO,
- DE = IO.
The first one is independent of the magnitude of angle C: DE is perpendicular to IO anyway. (This was observed some time ago by the participants of the Art of the Problem Solving forum. My insight was based on the realization that two additional lines, like DE, but constructed starting with the remaining sides of the triangle are parallel to DE and thus could not be perpendicular to IO because of the specific magnitude of the angles.) The second one is the consequence of the angle requirement.
The ingenious proof in Honsberger's book is by Hidetosi Fukagawa, the co-author with D. Pedoe of an important book Japanese Temple Geometry Problems, from which this site drew several examples and which appears to be responsible for much of the hype surrounding the Sangaku problems.
Extend AI to meet the circumcircle of ΔABC in F. The incenter being the meeting point of the angle bisectors of a triangle, AI bisects ∠A of the isosceles ΔBAE and thus also serves as the perpendicular bisector of the base BE. In particular,
Further, ΔAEF = ΔBEF by SAS or SSS, making
In ΔBEF, angle between equal sides EF and BF is 60°. Hence, ΔBEF is equilateral. So that
Next, we'll show that ΔBIF is isosceles:
|∠IBF||= ∠IBC + ∠IBF|
|= ∠B/2 + ∠A/2|
Importantly, we can now conclude that IF = BE. With this in hand, we can show that
Hence triangles IFO and BDE are equal by SAS, giving also IO = DE.
But then also ∠FIO = ∠BED. And since the two arms, IF and BE, of this angles are perpendicular, the other two arms, IO and DE, are also perpendicular.
- R. Honsberger, From Erdös To Kiev, MAA, 1996.
Copyright © 1996-2018 Alexander Bogomolny