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See complete Problem 40Triangle, Incenter, Excenter. Level: High School, SAT Prep, College geometryPost your solutions or ideas in the comments.
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)/dAjit: firstname.lastname@example.org
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)/dAjit: email@example.com
find a different answerhttp://geometri-problemleri.blogspot.com/2009/11/problem-40-ve-cozumu.htmlwhat do you think?
I propose a solution without sine rule, as follows.Take M on BC, with CM=b, so that BM=a-b, and, as ang(ACB)=60º, ACM is equilateral.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.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.Then we have AI=AD=DM=BM=a-b.BCE and ICA are similar, so BE/AI = BC/CI, DE/(a-b) = a/d and finally DE = a(a-d)/d.
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 only
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.