tag:blogger.com,1999:blog-6933544261975483399.post3471455681405196900..comments2024-02-22T05:31:28.964-08:00Comments on Go Geometry (Problem Solutions): Problem 21: Acute triangle, Orthocenter, DiameterAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-39753036571367799272015-01-16T13:42:18.226-08:002015-01-16T13:42:18.226-08:00MANEO is cyclic, AO its diameter. BFHE is cyclic, ...MANEO is cyclic, AO its diameter. BFHE is cyclic, so AF.AB=AH.AE ( 1 ), AM is tangent to the circle (BMC), and AM^2=AF.AB ( 2 ); with (1) AM^2=AH.AE and AM is tangent to the circle (HEM), so <AMH=<MEA ( 3 ). In a similar way <ANH=<NEA ( 4 ). Since <MAN+<MEN = 180, H belongs to the line MN. Best regards, Stan FulgerAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-86482507519290679742012-09-28T21:44:15.331-07:002012-09-28T21:44:15.331-07:00pascal's theorem is concerned with projective ...pascal's theorem is concerned with projective geometry ....... how can i prove it through elementary geometry ?????? as it ishttps://www.blogger.com/profile/12504861338709342832noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-883636922329204162010-06-24T16:09:10.334-07:002010-06-24T16:09:10.334-07:00Let DF and BC intersect at P. Let the intersection...Let DF and BC intersect at P. Let the intersection of the tangents at F and B be X, and let the intersection of the tangents at D and C be Y. FC and BD intersect at H, so by Pascal's theorem on FFCBBD and DDFCCB, X,Y,H, and P are collinear. The polars of X and Y are BF and CD and intersect at A. Since X, Y, and H are collinear, the polar of H must concur with the polars of X and Y, that is, the polar of H must pass through A. It follows that H lies on the polar of A, which is line MN, so we are done.zherohttps://www.blogger.com/profile/11757325056346633732noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-53141650263523018662010-01-02T02:23:37.816-08:002010-01-02T02:23:37.816-08:00the circles of diameter BC and of diameter AO
inte...the circles of diameter BC and of diameter AO<br />intersect at the tangency points M,N.<br />the line MN is their radical line<br />AOE is a right triangle,E lies on the circle of <br />diameter AO<br />the circle power of H with respect to this circle<br />is P1(H)=-HA.HE<br />BDC is a rigth triangle,D lies on the circle of diameter BC<br />the circle power of H with respect of this circle<br />is P2(H)= -HB.HD<br />but HA.HE=HB.HD =4R²cosAcosBcosC<br />H lies on the radical line,<br />M,H,N are collinear<br />.-.Anonymousnoreply@blogger.com