tag:blogger.com,1999:blog-6933544261975483399.post9029050696813916722..comments2024-05-21T05:01:49.873-07:00Comments on Go Geometry (Problem Solutions): Elearn Geometry Problem 40Antonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-6933544261975483399.post-23543555109945576562012-04-01T07:20:05.895-07:002012-04-01T07:20:05.895-07:00Thanks Peter for your comment. Also, 'b' c...Thanks Peter for your comment. Also, 'b' can be calculated in terms of 'a' only, and ED can be calculated in terms of 'a' only.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-25867202830110438172012-03-31T22:50:23.672-07:002012-03-31T22:50:23.672-07:00Please note that values of a,b and d are not indep...Please note that values of a,b and d are not independent . With given angles B and C we can calculate value of d in terms of a and b. In this problem ED can be calculated in terms of a and b onlyPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-61122546932342143542012-03-11T09:12:58.201-07:002012-03-11T09:12:58.201-07:00I propose a solution without sine rule, as follows...I propose a solution without sine rule, as follows.<br />Take M on BC, with CM=b, so that BM=a-b, and, as ang(ACB)=60º, ACM is equilateral.<br />Ang(BDE)=70º, because it is external to BDC, and also ang(DBE)=70º, because EB is bissector of the external angle on B. Thus EB=ED.<br />The bissector CN and the side AM are perpendicular, ADM is isosceles, and ADN and MDN are congruent. Also ADN and AIN are congruent, so AI=AD=DM. Furthermore, ang(ADC)=ang(CDM)=70º and ang(BDM)=40º. Thus DM=BM.<br />Then we have AI=AD=DM=BM=a-b.<br />BCE and ICA are similar, so BE/AI = BC/CI, DE/(a-b) = a/d and finally DE = a(a-d)/d.Nilton Lapanoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-50640269546025100552009-04-30T21:36:00.000-07:002009-04-30T21:36:00.000-07:00In my solution above the last line should've r...In my solution above the last line should've read: Equations (1) & (2) give us, (a-b)/d = x/a or x=DE=a(a-b)/d<br />Ajit: ajitathle@gmail.comAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-21193495137566645032009-04-28T22:32:00.000-07:002009-04-28T22:32:00.000-07:00Here's a solution which uses only the sine rul...Here's a solution which uses only the sine rule and some construction. Join E to B and locate a point M on BC such that CM=CA=b. Thus, BM = a-b. Since CA=CM, angle CAM=angle AMC. But since angle C=60, we've angle CAM=angle AMC =60 which makes angle BMA=120 and angle BAM=20. Hence, (a-b)/b = sin(20)/sin(40). Now join A to I. From triangle IAC, d/b = sin(40)/sin(70). Hence, (a-b)/d = sin(20)*sin(70)/sin(40)*sin(40) = sin(70)/2cos(20)sin(40) = 1/2sin(40) --(1). Now since EB is external bisector of angle B we've angle EBD=70 while angle BDE=40+30=70. Thus, EB=ED=x. Triangle EBC gives, x/a =sin(30)/sin(40) = 1/2sin(40) ---(2). Equations (1) & (2) give us, (a-b)/b = x/a or x=DE=a(a-b)/d<br />Ajit: ajitathle@gmail.comAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.com