tag:blogger.com,1999:blog-6933544261975483399.post2928756006817117437..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1054: Triangle, Altitude, Orthocenter, Circumcenter, Perpendicular, MidpointsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-33588950456044324162021-04-22T11:11:45.738-07:002021-04-22T11:11:45.738-07:00Clearly O is center of 9 pt circle of ABC. Let P b...Clearly O is center of 9 pt circle of ABC. Let P be mid-point of BH and Q foot of BH on AC. P,Q lie on the same circle. <br />Since m(PQF)=90=> P,O,F are collinear <br />PM // AB and m(PMF)=90 => MF // HN and similarly NF//MH => HMFN is parallelogram and G its centroid => HGF is one of its diagonal <br />Since G mid-point of HF and OG // PH => OG=PH/2=BH/4Sailendra Thttps://www.blogger.com/profile/12056621729673423024noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-13527262153311001212014-11-02T03:19:08.255-08:002014-11-02T03:19:08.255-08:00Let BH intersect circle O (9-pt circle of ΔABC) at...Let BH intersect circle O (9-pt circle of ΔABC) at L.<br />First observe Δs LMN, FMN have same circumcircle.<br />L bisects BH. So LM // BA , so perp to HN. already BH perp MN.<br />So H is the orthocenter of ΔLMN.<br />Hence BH = 2 LH = 2(2 OG) = 4 OG..Pravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-58925930934486406662014-11-01T22:36:35.225-07:002014-11-01T22:36:35.225-07:00http://s25.postimg.org/gymbfhbjz/pro_1054.png
Not...http://s25.postimg.org/gymbfhbjz/pro_1054.png<br /><br />Note that circumcircle of triangle MNF is 9 points circle of triangle ABC<br />Since N and F are midpoints of HC and AC => NF=1/2 AH=MH<br />And NF//MH => HNFM is a parallelogram<br />Since G is the midpoint of MN => H, G, F are collinear and GF=GH<br />Since ∠(LKF)= 90 degree => FOL is a diameter of circle O => GO= ½ GL= ¼ HB<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-48864179895620148132014-11-01T18:31:48.720-07:002014-11-01T18:31:48.720-07:00Obviously H is the orthocenter of ΔABC, and circle...Obviously H is the orthocenter of ΔABC, and circle O is its nine-point circle. <br /><br />Let P be the circimcenter of ΔABC, then BH=2×PF. <br />Now H,O,P are the Euler Line with HO=OP. <br />Also clearly, H,G,F are the median of ΔAHC, with HG=GF. <br /><br />Thus, PF=2×OG, and hence BH=2×PF=4×OG. Jacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.com