tag:blogger.com,1999:blog-6933544261975483399.post7803430658996584761..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1284 Two Equilateral Triangle, Midpoint, CongruenceAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-37550268003410688082016-11-08T13:06:26.024-08:002016-11-08T13:06:26.024-08:00Dear Antonio
My solution does not appear yet, hen...Dear Antonio<br /><br />My solution does not appear yet, hence resenting.<br /><br />Complete the rhombus ABCX.<br /><br />Triangles ACE ≡ XCD, SAS.<br /><br />So AE = XD and hence FH = FG from the mid point theorem in triangles ACE<br />and XBD.<br /><br />Similarly we can show that GH = FH by completing rhombus CDEY.<br /><br />Hence FG = FH = GH<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-11997264063696297412016-11-08T03:13:45.375-08:002016-11-08T03:13:45.375-08:00Complete the rhombus ABCX.
Triangles ACE ≡ XCD, S...Complete the rhombus ABCX.<br /><br />Triangles ACE ≡ XCD, SAS.<br /><br />So AE = XD and hence FH = FG from the mid point theorem in triangles ACE<br />and XBD.<br /><br />Similarly we can show that GH = FH by completing rhombus CDEY.<br /><br />Hence FG = FH = GH<br /><br />Sumith Peiris<br />MoratuwaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-68724341689395546842016-11-07T09:46:16.736-08:002016-11-07T09:46:16.736-08:00Problem 1284
Let the equilateral triangle AEP ...Problem 1284<br />Let the equilateral triangle AEP (PA,PE intersects the BD, not the extensions ). Then triangle ABP=triangle ACE (AB=AC,AP=AE, <BAP=60-<PAC=<CAE).So BP=CE=CD,but<br /><PBC+<BCD=<ABP-60+<BCD=<ACE-60+<BCD=240-60=180, then BP//CD so the BPDC is<br />Parallelogram .Then the points P,G and C are collinear with PG=GC.So GF=AP/2=AE/2=FH<br />and <HFG=<EAP=60 (GF//AP,FH//AE).Therefore triangle FGH is equilateral.<br />APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE<br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-39514820682946943972016-11-07T07:08:08.612-08:002016-11-07T07:08:08.612-08:00https://goo.gl/photos/49eNJ5Yuiq4nrZ5P8
Draw M on ...https://goo.gl/photos/49eNJ5Yuiq4nrZ5P8<br />Draw M on BF such that FM=FB ( see sketch)<br />We have triangle ACM is equilateral<br />Triangle ACE congruent to MCD ( case SAS)<br />And triangle MCD is the image of ACE in the rotational transformation center at C, rot. Angle= 60 degrees.<br />So AE=MD and angle(AE, MD)= 60 <br />And FH=1/2.AE and FH//AE<br />FG=1/2.MD and FG//MD<br />So FG=GH and angle(FG, FH)= angle (AE,MD)= 60 degrees<br />And FGH is equilateral<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com