## Monday, May 19, 2008

### Elearn Geometry Problem 40

See complete Problem 40
Triangle, Incenter, Excenter. Level: High School, SAT Prep, College geometry

1. 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
Ajit: ajitathle@gmail.com

2. 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
Ajit: ajitathle@gmail.com

3. 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.