tag:blogger.com,1999:blog-6933544261975483399.post7223939644616396420..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1374: Isosceles Triangle, Exterior Cevian, Incircle, Excircle, Tangency Points, Parallel LinesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-10026160980660777882018-08-14T06:52:48.089-07:002018-08-14T06:52:48.089-07:00Drop a Perpendicular from B to AC and meet at P
Dr...Drop a Perpendicular from B to AC and meet at P<br />Drop perpendicular from E to BP and meet at Q <br />Join BE and let m(ABE)=m(EBF)=@ and m(BAC)=m(BCA)=$ => m(GBC)=$-@ -------------(1)<br />Connect QH and form the right tr. QPH<br />In Right tr.QPH, QP=EF (radius of small circle) ------(2)<br />PH=BF------------(3)<br />Hence QPH and EFB are congruent (SAS) <br />=>m(QHP)=@ ---------------(4)<br /><br />Since m(EFB)=m(EQB)=90 => BFEQ are concylic <br />=>m(FBE)=m(FQE)=@ ---------------(5)<br /><br />From (4) and (5), F,Q and H are collinear <br />Simple angle chasing in tr.BDA gives m(BDA)=m(FDH)=$-2@ <br />Hence, m(HFD)=180+@-$------------------(5)<br />We know m(GBD)=180-m(GBC)=180-$+@---------------(6)<br /><br />From (5) and (6) BG//HFSailendra Thttps://www.blogger.com/profile/12056621729673423024noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-91028104434162696182018-08-06T09:17:35.492-07:002018-08-06T09:17:35.492-07:00Let the altitude BKL of Tr. ABC meet AC at L and t...Let the altitude BKL of Tr. ABC meet AC at L and the parallel to AC thro E at K.<br /><br />Let BL = h, AB = BC = a, BF = a - b, CH = c, AL = LC = d<br /><br />Let < FBE = < FKE = x, BFEK being cyclic<br />Let < KHL = y<br /><br />As per my solution to Problem 1373, BK = h - r1 = r2 = GH<br />So BKHG is a parallelogram and so KH //BG<br /><br />Now tan x = r1/(a-b) and tan y = r1/(c + d)<br /><br />But since a - b = c + d again as per problem 1373, tan x = tan y<br />Hence x = y<br />So FKH must be collinear since EK // AH<br /><br />So finally FKH // BG<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.com