tag:blogger.com,1999:blog-6933544261975483399.post6405625282119866925..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1295 Right Triangle, Incenter, Incircle, Excenter, Excircle, Congruence, AngleAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-6933544261975483399.post-71048343984290381302019-02-06T09:34:00.313-08:002019-02-06T09:34:00.313-08:00It is not clear why FG= FC . please explainIt is not clear why FG= FC . please explainPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-86979204478890017322018-10-18T07:28:43.193-07:002018-10-18T07:28:43.193-07:00BICE are concyclic with IE being the diameter. Let...BICE are concyclic with IE being the diameter. Let F be the center of this circle<br />Let G be the circumcenter of ABC and is the mid-point of AC<br />Since AC=IE => the above two circles are congruent with BC being the common chord<br />Connect FG and observe that FG perpendicular to BC and FG=FC=GC=Radius of the circles<br />=> m(FGC)=A=60<br />=> m(ACB)=30Sailendra Thttps://www.blogger.com/profile/12056621729673423024noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-31364948453030357132016-12-20T11:53:25.865-08:002016-12-20T11:53:25.865-08:00Let M,N be the contact points of incircle, excircl...Let M,N be the contact points of incircle, excircle respectively with AB; easily MN=BC, so BC is projection of AC onto BC, MN projection of IE onto AB. Since EI=AC, <A=2<C, thus <C=30.<br /><br />Best regardsStan Fulgerhttps://www.blogger.com/profile/07420210614111479737noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-88185059029148847052016-12-18T20:27:45.423-08:002016-12-18T20:27:45.423-08:00Problem 1295
In-radius r = ½(c+a-b) and ex-radi...Problem 1295<br /> In-radius r = ½(c+a-b) and ex-radius ra = ½(b+c-a)<br /><br /> (ra+r)2 + (ra-r)2 = b2<br /> 2(ra2 + r2) = b2<br /> (c+a-b)2 + (b+a-c)2 = 2b2<br /> Which simplifies to a2+c2 =2bc<br /> So b2 = 2bc and so b =2c<br /><br /> Hence ABC is a 30-60-90 triangle and <C = 30<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-62337239868461880412016-12-17T14:41:42.252-08:002016-12-17T14:41:42.252-08:00Draw from I parallel to AB, from E parallel to BC ...Draw from I parallel to AB, from E parallel to BC they meet at K<br />Tr IEK congr to Tr ABC => ang EIK = ang C, but ang EIK = 1/2 ang A<br />=> ang C = 30 <br />c.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-69648794743470165522016-12-17T12:06:35.944-08:002016-12-17T12:06:35.944-08:00https://goo.gl/photos/cbTJjt1P5rtiULgY9
Draw poin...https://goo.gl/photos/cbTJjt1P5rtiULgY9<br /><br />Draw points F, N, P, Q and M per attached sketch<br />Observe that BN=CP=CQ=IF= inradius of triangle ABC<br />Triangle CFI congruent to EQC ( case ASA)<br />So CI=CE => ICE is isoceles right triangle => CM/IE= ½<br />In right triangle ACM , CM/AC= CM/IE= ½<br />So ACM is 30-60-90 triangle <br />And angle A= 60 and angle C= 30<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com