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 657.
See problem # 328. J & K are midpoints of AB & AC resply. Hence JK = BC/2 =(BE+CE)/2=(BD+CF)/2 = 4.5Ajit
http://img143.imageshack.us/img143/1590/problem657.pngConnect DF and AE1. We have (DEF)=(ADF) both angles look the same arc DF but (AHF) supplement to (DEF)So (AHF) supplement to ( ADF) >> A,D,F, H is cyclicSimilarly we also have A,G,D,F cyclic 5 points A,G,D,F,H are cyclic ( see picture)2. We have (EDF)= (FEC) both angles look the same arc EF And ( EDF)=(GHF) both angles look the same arc GFSo ( FEC)=(GHF) and GH //BC3. Parallelogram AGEH have diagonals bisect each other so AJ/AB= ½ and KJ= 4.5Peter Tran
Let JK meet AE at LSince AGEH is a parellelogram AL = LE. But LK//EC so from mid point theorem LK = EC/2 = 2Similarly JL = BE /2 = 2.5So JK = 4.5Sumith PeirisMoratuwaSri Lanka