tag:blogger.com,1999:blog-6933544261975483399.post3205080992634265188..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 828: Quadrilateral, Midpoint, Opposite sides, Diagonals, Concurrent lines, Triangle, BimedianAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-6933544261975483399.post-45436360582488391652013-01-18T02:14:37.235-08:002013-01-18T02:14:37.235-08:00Let A,B,C,D are points of mass 1.
Since E,F,G,H,...Let A,B,C,D are points of mass 1. <br /><br />Since E,F,G,H,M,N are four mid-points, we have<br />E = 1/2(A+B)<br />F = 1/2(B+C)<br />G = 1/2(C+D)<br />H = 1/2(D+A)<br />M = 1/2(A+C)<br />N = 1/2(B+D)<br /><br />Mid-point of EG = 1/2(E+G) = 1/4(A+B+C+D)<br />Mid-point of FH = 1/2(F+H) = 1/4(A+B+C+D)<br />Mid-point of MN = 1/2(M+N) = 1/4(A+B+C+D)<br /><br />Let O=1/4(A+B+C+D)<br />Then EG, FH, MN are concurrent at their common mid-point O. Jacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-13168016979069405102012-12-02T13:34:12.407-08:002012-12-02T13:34:12.407-08:00http://www.youtube.com/watch?v=_YzAMXsARSQhttp://www.youtube.com/watch?v=_YzAMXsARSQAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-91357814816394704882012-12-02T11:31:26.078-08:002012-12-02T11:31:26.078-08:00FM es base media en tr ABC así que FM // AB.
NH es...FM es base media en tr ABC así que FM // AB.<br />NH es base media en tr ADB así que HN // AB.<br />Finalmente GM // HN.<br />Análogamente FN // MH.<br />Entonces MFNH es paralelógramo y O es el pto medio de una diagonal, luego O pertenece a MN y OM=ON.Editorhttps://www.blogger.com/profile/18079120609888942700noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-44601556428922956272012-11-29T21:50:52.314-08:002012-11-29T21:50:52.314-08:00[Vector Method]
Let Vector(BA) = u ; Vector(BC) = ...[Vector Method]<br />Let Vector(BA) = u ; Vector(BC) = v ; Vector (BD) = ru + sv , where r,s are scalars.<br />Hence <br />BE = (1/2)u ;<br />BF = (1/2)v ;<br />BH = (1/2)(BA + BD) = (1/2)[(r+1)u + sv]<br />BG = (1/2)(BC + BD) = (1/2)[ru + (s+1)v]<br />BM = (1/2)(u + v)<br />BN = (1/2)(ru + sv)<br /><br />Now The mid point of MN = (1/2)*(BM + BN) = (1/2)*(1/2)*[(r+1)u + (s+1)v]<br />This can be rewritten into (1/2)*{(1/2)*u + (1/2)[ru + (s+1)v]} = (1/2)*(BE+BG) <br />i.e. the mid point of EG.<br />Symmetrically, it would also be the mid point of FH.<br /><br />Q.E.D.W Fungnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-36547674835905942112012-11-29T15:57:40.796-08:002012-11-29T15:57:40.796-08:00Connect EF,FG, GH and EH
We have FG//BD//EH and FG...Connect EF,FG, GH and EH<br />We have FG//BD//EH and FG=1/2. BD=EH<br />So EFGH is a parallelogram so EG and FH intersect at midpoint of diagonal.<br />Connect MF,FN,NH and MH<br />We have FN//CD//MH and FN=1/2.CD=MH<br />So MFNH is a parallelogram so diagonals MN and FH intersect at midpoint of diagonal.<br />So EG, FH and MN are concurrent at the midpoint O<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com