Posted in geometry, Homeschooling, Simple math by: admin
No Comments
01 Feb
The three angle bisectors of a triangle meet at incenter of the triangle. Reversing the problem we may ask a relevant question:

Given three concurrent lines: α, β, and γ. Is there always a triangle with the three lines as the angle bisectors. If so, construct the triangle.
Solution
Given three concurrent lines: α, β, and γ. Is there always a triangle with the three lines as the angle bisectors. If so, construct the triangle.

Solution
The answer to the question is in positive, and the triangle can be constructed is as follows.

First construct a triangle for which the three lines serve as the altitudes. The orthic triangle of the latter is the one we are after, because of the mirror property the orthic triangles possess.
Obvioiusly, the construction is not unique, what is unique is the shape of the triangles - they all are similar.
Related Posts
Geometry in ProvidenceProvidence, RI is an enchanting city. Here are three pictures from a recent trip. A frieze pattern. Triangles - to color and to count. Vanity of human travails.
Beautiful GeometryAnd this is how it goes: 51 chapters that combine pedagogically meaningful artwork together with informative, and often eye opening, text. The book ends with a short Appendix which lays foundations for several mathematical concepts mentioned in the text.
This is truly an enjoyable, simple book that meets if not exceeds the author's expectations. It's a good seasonal present, too.
A square in parallel linesThe problem is credited to V. V. Proizvolov and may serve an example where using dynamical software to sketch a diagram proves to be a distraction. I put together a GeoGebra applet and looked at the possible properties of the configuration with several elements added. Meanwhile, Hubert Shutrick pointed out its obvious, salient feature. The solution is a one liner