tag:blogger.com,1999:blog-6933544261975483399.post4875366678204713302..comments2022-09-27T03:11:10.165-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1207: Triangle, Circle, Incenter, Circumcenter, Excenter, Circumradius, PerpendicularAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-5975811753077694462016-04-15T13:53:42.751-07:002016-04-15T13:53:42.751-07:00Bring IK⊥AC,MN⊥AC, EL⊥AC(M is midpoint arc AC).Is...Bring IK⊥AC,MN⊥AC, EL⊥AC(M is midpoint arc AC).Is &lt;MIC=&lt;ABC/2+&lt;ACB/2 , &lt;MCI=&lt;MCA+&lt;ACI=&lt;ABC/2+&lt;ACB/2.Therefore &lt;MIC=&lt;MCI.So MI=MC.But IC⊥CE.<br />Then &lt;MIC+&lt;MEC=90 and &lt;MCI+&lt;MCE =90 therefore MI=MC=ME.So MO//ED<br />Is IO=OD. Therefore ED=//2OM=2R.<br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-11564035850722563952016-04-14T23:25:38.587-07:002016-04-14T23:25:38.587-07:00Join BIE. Let it cut circle(O)at M. Angle ECI is a...Join BIE. Let it cut circle(O)at M. Angle ECI is a right angle.<br />M is the midpoint of arc AMC. So OM bisects AC at right angles.<br />We are done if we can show O is the midpoint of ID.<br />Let X, Y, Z be the projections of I. O. D on AC respectively.<br />It is easy to see that XY = b/2 - (s -a) = (c -a)/2 = YZ.<br />IX, OY, DZ being //, OYM is a midline // to DE in Triangle IDE.<br />Hence DE = 2 OM = 2R.<br />N Vijaya Prasad<br />Rajahmundry - INDIA.Pravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-33518176506569473942016-04-14T21:01:47.203-07:002016-04-14T21:01:47.203-07:00Lets assume touch point of incircle and excircle a...Lets assume touch point of incircle and excircle are F and G, and Midpoint of AC is M. It is easy to see that FM=GM=(a-c)/2 Also IF,OM and DG are parallel to each other ( all are perpendicular to AC), hence O is midpoint of ID.<br />B, I and E are collinear, join BE and let it intersects circumcircle at point H, and IH=HE, thus H is midpoint of IE.<br />Consider triangle IED, O is midpoint of ID and H is midpoint of IE,<br />Hence DE=2*OH, since OH=R, DE=2R. Pradyumna Agashehttps://www.blogger.com/profile/10300531209692781145noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-69607766509525577282016-04-14T17:42:18.008-07:002016-04-14T17:42:18.008-07:00http://s22.postimg.org/nia3yt2fl/pro_1207.png Let...http://s22.postimg.org/nia3yt2fl/pro_1207.png<br /><br />Let L, M, N , P are points as shown on the sketch<br />Observe that AL= NC= half perimeter of triangle ABC- BC<br />P is the midpoint of arc AC =&gt; M is the midpoint of AC and LN<br />In trapezoid ILND , MO is the mid-base =&gt; O is the midpoint of ID<br />Triangle IOP similar to triangle IDE ..( case AA)<br />Since O is the midpoint of ID so DE= 2 x OP= 2.R<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com