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