Same is true if we use Brocard axis of triangles instead of Euler lines.
https://bernard-gibert.pagesperso-orange.fr/Exemples/k001.html, Property 20.
For P=X(13), OAOBOC is equilateral. It's the reflection of Napoleon Equilateral triangle on its center.
https://www.journal-1.eu/2021/Abdilkadir%20Altintas.%20Congruent%20Circles%20on%20Locus%20Problems,%20pp.%2092-96..pdf
Let H=X(4)-Orthocenter of ABC. A 1 B 1 C 1 cevian triangle of H. B1 A , C1 A are reflections of B1, C1 on A. L A , line through B1 A , C1...
Same is true if we use Brocard axis of triangles instead of Euler lines.
ReplyDeletehttps://bernard-gibert.pagesperso-orange.fr/Exemples/k001.html, Property 20.
ReplyDeleteFor P=X(13), OAOBOC is equilateral. It's the reflection of Napoleon Equilateral triangle on its center.
ReplyDeletehttps://www.journal-1.eu/2021/Abdilkadir%20Altintas.%20Congruent%20Circles%20on%20Locus%20Problems,%20pp.%2092-96..pdf
ReplyDelete