## Saturday, November 30, 2019

### Dynamic Geometry 1450: Ortholine, Steiner Line

Interactive step-by-step animation using GeoGebra. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

#### 1 comment:

1. https://photos.app.goo.gl/icHy7rRP4DjBxi6UA

Draw circumcircles of triangles ADE, ABF, BEC and CDF
Per the result of problem 547 all these 4 circumcircles will concur at a point G ( see sketch)
From G make perpendicular projections to AF, BF ,ED and AE at M, N, L, P.
Observe that MNL, NLP, MNP and MLP are Simson lines of G to circumcircles of triangles DCF, EBC, ABF and AED.
So M, N, L and P are collinear.
Perform homothety transformation center G with scale factor= 2 , MNLP will become M’N’L’P’. and M’N’L’P’ are collinear
M’N’L’ will be Steiner line of triangle CDF from point G .
This line will pass through orthocenter H3 of triangle CDF ( property of Steiner line)
Similarly for other 3 triangles , we will have Steiner lines N’L’P’, M’N’P’ and M’L’P’ from G of triangle BEC, ABF and AED.
M’N’L’P’ will pass through Orthocenters H1, H2, H3, and H4 => H1, H2, H3 and H4 are collinear