tag:blogger.com,1999:blog-6933544261975483399.post3491440589552948852..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 506: Triangle with Three Squares, Center, Midpoint, Perpendicular, CongruenceAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-89447091830627678262014-07-12T17:50:54.049-07:002014-07-12T17:50:54.049-07:00From problem 496, AH and DC meet orthogonally at p...From problem 496, AH and DC meet orthogonally at point P on line EC. <br />Let M1 be intersection of circumcircle of APC with line EC, which must be proven to coincide with M. <br />Also from 496, <EPA=<GPC=45.<br />Using cyclic condition, <M1CA=<M1PA=45 and <M1AC=<GPC=45 making AM1C isosceles right triangle.<br />AH=CD, AM1=CM1, and <M1AH=<M1CD from cyclic quadrilateral ACPM1, so triangle M1CD is 90 rotation of triangle M1AH about M1.<br />Triangle EAM1 is reduced rotation of triangle DAC by factor of sqrt(2). Same argument applies for triangles GCM1 and HCA. However, since DC=AH, EM1=GM1=DC/sqrt(2).Ivan Bazarovnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-10392419396341621892010-08-20T12:57:20.743-07:002010-08-20T12:57:20.743-07:00Aplicamos el teorema de Botema en el segmento AC y...Aplicamos el teorema de Botema en el segmento AC y obtenemos que AM=MC, que AC=2MX (donde X es proyeccion ortogonal de M en AC) y que <AMC=90.<br />Como AO=OC y <AOC=90 tenemos que AOCM es cuadrado.<br />by mathreyesAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-76501011664660802752010-08-19T22:43:56.160-07:002010-08-19T22:43:56.160-07:00Let Q is the projection of E on AC ; N is the proj...Let Q is the projection of E on AC ; N is the projection of M on AC<br />Let K is the projection of B on AC; P is the projection of G on AC<br />We will prove that N is the midpoint of AC and MN=1/2* AC<br />Since M is the midpoint of EQ so N is the midpoint of PQ .<br />We have CP=CQ.cos(90-C)= a.sin(C) = BK<br />GP=a.cos(C )= CK<br />QA= c.cos(A) = BK and EQ=c.cos(A)= AK<br />Since CP=QA so N will be midpoint of AC.<br />In the trapezoid EQPG we have MN=.5*(EQ+GP)=.5(AK+CK)=.5*AC<br />So AMC is a right and isosceles triangle and AMCO is a square.<br />Similarly if we project points E, M, B, G over DH and with the same way as above<br />We can show that DMH is a right and isosceles triangle . <br /><br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com