Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Wednesday, May 21, 2014
Math Geometry Problem 1017: Square, Center, Midpoints, Circle, Concyclic Points, Cyclic Quadrilateral
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∠ECD = ∠EFD = ∠EGD = ∠EHD = 90° because of square symmetry. Hence C,E,F,G,H,D are concyclic. ED is a diameter. So is CH by symmetry.
ReplyDeleteProblem 1017
DeleteTo Anonymous Why angle EGD = 90 degree?
http://s2.postimg.org/r4m3ldq6x/pro_1017.png
ReplyDeleteDraw points L, I, L as per attached sketch
Note that triangles ILC congruent to triangle LFI… (case SAS)
So IC=IF=IG=IE=ID=IH => so all these points are concyclic
If you draw a 4-by-4 grid, GE is a 1x3 move and GD is 3x-1 move, so ∠ECD=90°
ReplyDeleteECDH es un rectángulo, por lo tanto es inscriptible en una circunferencia C1.
ReplyDeleteEFGH es un trapecio isósceles, por lo tanto es inscriptible en un circunferencia C2.
Como el rectángulo y el trapecio tienen el segmento EH en común C1=C2.
Entonces C, D, E, F, G y H pertenecen a la misma circunferencia, son concíclicos.
To Anonymous Problem 1017
ReplyDeletePlease, send your solution in Spanish to
(Enviar su solucion en Espanol a:)
Problema de Geometría 1017
Thanks