Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 976.
Let D be a point on BC such that AD bisects ∠A. Then AID is straight line. ∠BID=∠BAI+∠ABI=67.5°∠BDI=90°−∠BAD=67.5°Thus, BI=BD. Hence, by Problem 975, AC=AB+BD=AB+BI.
http://imageshack.com/a/img716/8509/6hr3.pngOn BC locate E such that BI=BETriangle IBE is isosceles with ∠ (IED)=22.5 degrees= ∠ (ICD)Triangles IDC and IDE are congruent…. ( Case SAS) => DC=DEAnd CA= 2 x CD=CE => CA=CB+BE= AB+BI
The angle bisector BI divides isoceles right Tr. ABC. into 2 congruent isoceles right Tr. s. So if the in radius be r, AB - r = AC / 2 and BI + r = AC/2. Adding these 2 equations the result followsSumith PeirisMoratuwaSri Lanka
Problem 976Let point D in the extension of AB to B such that AD=AC.Then triangle ADI and ACI are congruent (AD=AC,AI=AI,<DAI=<CAI=22.5 ).So <BDI=<AC[=22.5 and <ABI=<BDI+<BID or45=22.5+<BID or <BID=22.5 therefore BI=BD and AC=AD=AB+BD=AC+BI.APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE
(More of an algebric solution)Let AB=BC=1 => AC = √2 and BI extended meet AC at PSince m(BPC) = 90, triangle BPC is isosceles right angled=> BP = CP = 1/√2Semi-perimeter of ABC , S=(2+√2)/2S.r=.5*AB.BC => r=IP=1/(2+√2)=> BI = BP-IP = 1/√2 - 1/(2+√2)=> BI = 2/2+2√2=> BI = 1/1+√2 = √2-1=> BI = AC-AB=> AC = AB+BI
2r=a+b-csqrt(2)BI=AB + BC - Ac =2AB-sqrt(2)ABso sqrt(2) (BI+AB)=2AB=sqrt(2)ACso BI+AB=AC