Episode 3: Lines and Points

On Tuesday 28th January at 7.30 pm, Tree House Bookshop

We focus in this talk on various ‘centres’ of a triangle where three lines are concurrent (meet in a single point). Sometimes three of these centres themselves all lie in a straight line (are collinear). We shall explore some quite surprising ways in which these centres are not only collinear but they divide their common line in very simple ratios.

In particular we examine the case of four collinear centres dividing the so-called ‘Euler line’ in a very satisfying manner. Every triangle has such an Euler line. There will be many opportunities to test our results (at home) by drawing so, if you are inclined, it will be worth finding an old geometry set, or investing in a new one (a compass at least). For those who prefer computer-based drawing, or are just happy to watch, we are planning to demonstrate the theory by means of software tools (such as Geogebra).

The ‘starter’ slides (Setting the Scene) this time are part of the whole package.

[There was an error on slide 8 of the handout, and in main slides (‘Similar, not congruent’) – corrected here 29th January.]

Much of the ‘talk’ this time will be demonstrations and proofs, so not on the slides.

Detailed proofs for most of the results of concurrency and collinearity demonstrated in the session were not given. The links below give a .pdf (2 pages) with detailed proof of each result. The EulerLine proof includes the nine point circle diagram. Please let me know errors spotted or questions arising.

During discussion it was pointed out that Geogebra was available free of charge to individuals. The question was also raised about the accuracy of Geogebra. There is a bit more on this in the blog posts. A nice little ‘extra’, if you like that sort of thing, is to prove the fact I stated without proof at the end of the EulerLine pages … namely that the centre N of the nine point circle is indeed the midpoint of the Euler line OH.