tag:blogger.com,1999:blog-6933544261975483399.post4343472432087215714..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1281 Triangle, Circle, Diameter, Tangent, Altitude, Congruent AnglesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-36809029842447979682016-11-01T10:31:26.865-07:002016-11-01T10:31:26.865-07:00m(BOD) = mBOE) = m(BOH) = 90 => B,D,H,O & E...m(BOD) = mBOE) = m(BOH) = 90 => B,D,H,O & E are concylic <br />Since BD = BE => m(BHD) = m(BHE) = @ (say)<br />Therefore m(AHD) = m(CHE) = 90-@Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-21662155660221311702016-10-30T23:28:14.178-07:002016-10-30T23:28:14.178-07:00Problem 1281
Suppose that the DC intersect BH ...Problem 1281<br /><br />Suppose that the DC intersect BH in point P and AD the BH to Q.Then < DCA=<QDB ( cord and tangent) and <DQB=<DCA (perpendicular),so <DQB=<QDB or BD=BQ=BE. But <PDB=90-<BDQ=<DPB,then BD=BP=BE=BQ. So PE is perpendicular to BE.<br />Then at triangle PAC the point P is orthocenter, and A,P and E <br />Are collinear. The ADPH ,HPEC are concyclic. So <DHA=<DPA=<EPC=<EHC.<br />4 HIGH SCHOOL OF KORYDALLOS PIRAEUS-GREECE<br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-23003390159827569792016-10-30T21:55:33.675-07:002016-10-30T21:55:33.675-07:00Let AD, HB meet at P. Let HPB extended meet AD ext...Let AD, HB meet at P. Let HPB extended meet AD extended at X.<br /><br />Now P is the orthocenter of triangle AXC, hence AP is perpendicular to XC.<br />But AE is perpendicular to XC. It follows that A,P,E are collinear and XEC<br />are collinear.<br /><br />P being the orthocenter of triangle AXC, easily P is the incentre of<br />triangle HDE.<br /><br />So < AHD = < CHE<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-12732125302649735822016-10-30T18:09:42.963-07:002016-10-30T18:09:42.963-07:00https://goo.gl/photos/cZVha9LddjDaWn519
Connect O...https://goo.gl/photos/cZVha9LddjDaWn519<br /><br />Connect OB, OD ,OE where O is the midpoint of AC<br />We have OD⊥BD and OE⊥BE and OB is the angle bisector or angle BDE<br />So O, H, D, B, E are cocyclic<br />In circle OHDBE ∠AHD= ∠DBO<br />∠CHE=∠OHE= ∠OBE<br />But ∠DBO=∠OBE => ∠AHD=∠CHE<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com