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