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Hiroshi Haruki's Lemma: What is this about?
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The applet may suggest the following assertion:
Assume AB and CD are two chords and P a point on a circle. Let Q and R be the intersections of PC and PD with AB. Draw circumcircles of triangles PRC and PQD and let AB intersect the two circles in F and E, respectively. Then
| (1) | BE = AF, and their length does not depend on the position of P on the given circle. |
The latter is an unintended consequence of the proof of Hiroshi Haruki's Lemma [Honsberger]:
| (2) |
For AB, CD, P, Q, R as above, |
Proof
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In the circumcircle PQD, if P lies in the major arc CD, the angles QPD and QED are equal as subtending the same chord QD. As P traverses the major arc CD , the angle at E (AED) remains the same, which means that the point E remains fixed. (This is in fact true even if P lies on the minor arc CD.) So that
We apply the intersecting Chords Theorem to PD crossed by AE in two circles:
| PR·RD = QR·RE and PR·RD = AR·RB, |
from which
| QR·(RB + BE) = (AQ + QR)·RB, |
or QR·BE = AQ·RB, and finally
| (3) | AQ·RB / QR = BE. |
This proves (2) and also (1), since a similar argument in the circle PRC shows that F is independent of P and, in addition,
The lemma leads to elegant proofs of the Butterfly and Two Butterflies theorems.
References
- R. Honsberger, The Butterfly Problem and Other Delicacies from the Noble Art of Euclidean Geometry I, TYCMJ, 14 (1983), pp. 2-7.
- R. Honsberger, Mathematical Diamonds, MAA, 2003, pp. 136-140
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