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 778.
http://img213.imageshack.us/img213/5918/problem778.pngDraw lines per attached sketchWe have LC//BH and LB//CE So BLCH is a parallelogram and BH=LCIn right triangle ACL we have AC^2+CL^2=AL^2Or BH^2+AC^2=4.R^2
http://img232.imageshack.us/img232/3764/p778resuelto.png(look the construction of the picture).As AP is diameter, <PCA = 90° and <ABP=90°. Note that PC and HC are paralell to BH and BP respectively. Then BH = PC (because BPCH is paralelogram) and by pythagorean theorem we have that PC^2 + AC^2=4R^2. Hence BH^2+AC^2=4R^2
BH = 2R cos(ABC)...by Prob 777AC = 2R sin(ABC)...by Sine ruleBH^2 + AC^2 = 4R^2
By using the result in problem 777, the statement would be reduce to AC = 2R sinABC,which is obvious under sine law.
Let AO meet the circle at X. < AXB = < B so < BCE = 90. - B = < CAX = CBX hence BX//HC and since BF//XC BXCF is a parellelogram and it thus follows that BH = XCNow applying Pythagoras to right Tr. ACX the result followsSumith PeirisMoratuwaSri Lanka