tag:blogger.com,1999:blog-6933544261975483399.post3288661293643787377..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1408: Right Triangle, Incircle, Excircle, Incenter, Midpoint, Tangency Point, CollinearityAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-65158448295349232402018-12-15T01:09:18.745-08:002018-12-15T01:09:18.745-08:00If X & Y are the tangency points of the in cir...If X & Y are the tangency points of the in circle at BC & BA & if the radii of the incircle is p & of the excircle is q & S(ABC) = S, s the semi perimeter, <br /><br />XF/IX = (q-p)/p = (a-2p) / p = (a-2S/s)/(S/s)<br />= (as - ac)/(ac/2) = (s-c)/(c/2) = BF/BD<br /><br />So Tr.s BDF & XIF are similar and DIF are collinear <br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-44592722688649436782018-12-11T20:13:54.669-08:002018-12-11T20:13:54.669-08:00let incircle radius r, touch BC at H;
excircle rad...let incircle radius r, touch BC at H;<br />excircle radius R,<br />Right ABC, the length is a,b,c, c>a,c>b;<br />r=(a+b-c)/2; R=(c+a-b)/2;<br />if triangle IHF similar to triangle BDF, then D,I F colinear,<br />it easy to show r/(b/2) =(R-r)/R<br />so D,I,F colinear.<br /><br /> <br />Anonymousnoreply@blogger.com