Geometry Problem. Post your solution in the comments box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Thursday, December 3, 2015

### Geometry Problem 1169: Complete Quadrilateral, Orthocenters of the Component Triangles, Collinear Points, Ortholine, Steiner-line, Orthocentric line

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http://s18.postimg.org/sxwxdkf3d/pro_1169.png

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’.

M’N’L’ will be Steiner line of triangle CDF from point G .

This line will pass through orthocenter Hc 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 Ha, Hb, Hc, and Hd => Ha, Hb, Hc and Hd are collinear