tag:blogger.com,1999:blog-6933544261975483399.post7342100135813800311..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1283 Two Equilateral Triangle, Perpendicular, MidpointAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-68708601838699202342016-11-05T07:20:49.823-07:002016-11-05T07:20:49.823-07:00BECM is cyclic with BC as diameter so
< MEC = ...BECM is cyclic with BC as diameter so <br />< MEC = < MBC = 60<br /><br />ACDM is cyclic with AC as diameter so <br />< MDE < MAC = 60<br /><br />So 2 angles of Tr. MDE = 60 and is hence equilateral<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-29364776058208660152016-11-04T23:21:33.400-07:002016-11-04T23:21:33.400-07:00Problem 1283
Suppose that MN is perpendicular to...Problem 1283<br />Suppose that MN is perpendicular to CE ( DN =NE ,ABED is trapezoid ) so triangle MED is isosceles (ME=MD).But <AMC=90=<ADC then AMDC is cyclic. So <MDA=<MCA=60/2=30.<br />Then <MDE=90-30=60.Therefore triangle MDE is equilateral.<br />APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE<br /><br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-12051863622800531172016-11-04T16:03:21.137-07:002016-11-04T16:03:21.137-07:00https://goo.gl/photos/96rSC1H87PN4Xb7g7
Let N is ...https://goo.gl/photos/96rSC1H87PN4Xb7g7<br /><br />Let N is the projection of M over CDE<br />In triangle EMD, MN is the median and altitude from M => MED is isosceles triangle<br />Connect CM , CM ⊥AB and BE⊥EC => quadri. CMBE is cyclic<br />in cyclic quadrilateral CMBE ∠MDC=∠MBC=60 degrees => Triangle EMC is equilateral<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com