- Jul 4, 2014[Seiichi Kirikami]:
Dear friends, Dear Antreas,

(1) Given a triangle ABC and 3 squares whose diagonals coincide with BC, CA, and AB respectively, we denote the internal vertex of the square with BC-diagonal by D. Define E and F cyclically. We denote the intersection of BF and CE by A1. Define B1 and C1 cyclically.

AA1, BB1 and CC1 are concurrent at X(372). (sinA-cosA: : ).

(2) We denote the external vertex of the square with BC-diagonal by D1. Define E1 and F1 cyclically. We denote the intersection of BF1 and CE1 by A2. Define B2 and C2 cyclically.

AA2, BB2 and CC2 are concurrent at X(371). (sinA+cosA: : ).

P. S. This message is the continuation of message # 1747.

Best regards,

Seiichi.

_Dear Seiichi,Your constructions are simpler than the others mentionedin the respective points in ETC.We can generalize them:

Your BDC, CEA, AFB triangles are Kiepert triangleswith base angle = 45 d.We can take any angle omega (w) instead of 45 d.The perspectors (in barycentrics) are:1 / (cotA -+ cot(A-+w)) ::(the minus sign (-) for the internal case and the plus sign (+) for the external case)APH