tag:blogger.com,1999:blog-6933544261975483399.post5518353624744267649..comments2023-02-06T18:39:11.207-08:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1455: Nagel Point, Excircles, Incircle, Congruent SegmentsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-27580907454078198772020-03-02T21:09:50.719-08:002020-03-02T21:09:50.719-08:00https://photos.app.goo.gl/vEqkL8Da7ogdieyr5 Let a...https://photos.app.goo.gl/vEqkL8Da7ogdieyr5<br /><br />Let a, b, c are 3 sides of triangle ABC and 2p is its perimeter<br />Define point J, R(A) and r per attached sketch<br />From I(A) draw B’C’// BC<br />Note that AB’C’ is the image of ABC in the homothetic transformation<br />Since circle J is the excircle of triangle ABC =&gt; circle I is the excircle of triangle AB’C’<br />So AI(A)/ AT(A)= r/R(A)= (p-a)/p…..(1)<br />Apply Van Aubel II theorem in triangle ABC with point N<br /> we have AN/AT(A)= (p-b)/(p-a) + (p-c)/(p-a)= a/(p-a)<br />From above expression we will have NT(A)/AT(A)= (p-a)/p…..(2)<br />Compare (1) to (2) we have AI(A)=NT(A)<br />Similarly we will have same results with other 2 vertexes<br /> <br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-16951819563431046282020-03-01T02:22:20.779-08:002020-03-01T02:22:20.779-08:00Using Van Aubel&#39;s theorem We get (AN/NTa)=(AT...Using Van Aubel&#39;s theorem <br />We get (AN/NTa)=(ATc/TcB)+(ATb/TbC)=((s-b)/(s-a))+ ((s-c)/(s-a))<br />AN/NTa= a/(s-a), adding 1 to both sides we get ATa/NTa=s/(s-a)<br />Or NTa/ATa=(s-a)/s ........(1)<br /><br />If we draw a tangent to incircle which is parallel to BC, it will touch incircl at Ia. Lets assume altitude from A to BC is h. Then we get<br />AIa/ATa=(h-2r)/h, replacing h by 2S/a and r by S/s, we get <br />AIa/ATa= (s-a)/s .....(2) <br />Using eq (1) and (2) we have AIa= NTa <br />Similary BIb=NTb and CIc= NTc Pradyumna Agashehttps://www.blogger.com/profile/10300531209692781145noreply@blogger.com