A Property of the Line IO:
A Bundled Statement and Proof

R. Honsberger in [From Erdos to Kiev, pp. 199-201] treats a problem from Crux Mathematicorum, 1987, 160:

Now, the problem is a mixture of two:

1. DE is perpendicular to IO,
2. 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, BM = EM. Point F on the perpendicular bisector is then equidistant from B and E:

Further, ΔAEF = ΔBEF by SAS or SSS, making AFE = AFB, and implying BFE = 2AFB. But AFB = C on chord AB, and so

BFE = 2·30o = 60o.

In ΔBEF, angle between equal sides EF and BF is 60^o. Hence, ΔBEF is equilateral. So that

Next, we'll show that ΔBIF is isosceles: BF = IF. Indeed, AIB = 180^o - A/2 - B/2, so BIF = (A + B)/2. On the other hand, CBF = CAF (= A/2), as subtended by the same arc, so that

IBF = IBC + IBF   = B/2 + A/2   = BIF.

Importantly, we can now conclude that IF = BE. With this in hand, we can show that ΔIFO = ΔBDE. Since AI bisects A, F is the midpoint of the arc BFC, so that FO is perpendicular to BC. Additionally, IF is the perpendicular bisector of BE. It follows that the arms of angles IFO and EBD are respectively perpendicular and, therefore

IFO = DBE.

Now, central = 2C = 60^o, making ΔAOB equilateral, and we conclude that, surprisingly, the circumradius of ΔABC is just the side AB. Consequently,

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.

References

1. R. Honsberger, From Erdös To Kiev, MAA, 1996.

Copyright © 1996-2008 Alexander Bogomolny