tag:blogger.com,1999:blog-6933544261975483399.post7713011194828367233..comments2022-09-27T03:11:10.165-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1267 Triangle, Incircle, Excircle, Circle, Tangency Points, Perpendicular, 90 Degrees, ParallelogramAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-48597934963319019062016-09-29T04:18:33.907-07:002016-09-29T04:18:33.907-07:00With the usual notation, 2s = a+b+c, AE1= AF1 = C...With the usual notation, 2s = a+b+c,<br /><br />AE1= AF1 = CF =CD = s-c and<br />AE = AF = CF1 =CD1 = s-a<br /><br />From Problems 1265 and 1266 we have<br /><br />&lt;E1AG = &lt; G1CD1 and &lt; EAG = &lt; DCG, so &lt; GAG1 = GCG1<br />Further AG1/CG1 = AF1/CF1 = (s-c)/(s-a) = CF/AF<br /><br />Hence Tr.s GAG1 and GCG1 are similar. However the side GG1 is common to<br />these 2 triangles. Hence they must also be congruent and therefore AGCG1 is a<br />parallelogram.<br /><br />Hence Tr.s AF1G1 ≡ CGF (SAS), <br />so AG1 = GC<br />Similarly Tr.s AF1G ≡ CFG1 (SAS), <br />so AG = CG1<br /><br />So in quadrilateral FGF1G1, the opposite sides are equal, hence the same<br />is a parallelogram.<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-13634078212696246552016-09-29T03:06:08.550-07:002016-09-29T03:06:08.550-07:00Problem 1267 Is AF_1=CF(O=incircle, O_1=excircle ...Problem 1267<br />Is AF_1=CF(O=incircle, O_1=excircle corresponding to AC) or AE_1=AF_1=CF=CD. But triangleAE_1G_1 is similar with triangle G_1D_1C then E_1G_1/G_1D_1=AE_1/CD_1=DC/CD_1 so than vice versa Thales&#39; theorem we have<br />E_1D//G_1C. Also triangle AEG is similar with triangle GDC so EG/GD=AE/DC=AE/AE_1<br />then AG//DE_1.Terefore AG//CG_1.Similar CG//AG_1 so AG_1CG is parallelogram.<br />If AC Intersects GG_1 in M then G_1M=MG and F_1M=MF (AF_1=FC) so F_1G_1FG is<br />parallelogram.<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-90723360002758601492016-09-29T00:39:53.036-07:002016-09-29T00:39:53.036-07:00https://goo.gl/photos/1anav7EpeN3Uvv8e9 Let ED me...https://goo.gl/photos/1anav7EpeN3Uvv8e9<br /><br />Let ED meet AC at P and E1D1 meet AC at P1<br />Note that triangles BED and BE1D1 are isosceles and ED//E1D1<br />CF=AF1= p- AB where p= semi perimeter of triangle ABC<br />1. Per the result of problem 1265 and 1266 we have <br />FC/FA= PC/PA= k<br />We can calculate FP in term of k and AC <br />FP= ( (k/1+k) + k/(1-k)). AC <br />Similarly F1A/F1C=P1A/P1C= k and F1P1= ( (k/1+k) + k/(1-k)). AC<br />So FP=F1P1<br />Triangle PGF congruent to P1G1F1 ( case ASA) =&gt; FG=F1G1<br />We also have FG//F1G1 =&gt; FGF1G1 is a parallelogram<br /><br />2. we have triangle AFG congruent to CF1G1 ( case SAS) and triangle CGF1 congruent to AG1F1 ( case SAS)<br />So AG= CG1 and CG= AG1 <br />And AGCG1 is a parallelogram<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com