Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem.Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 1000.
CD=AE=2GH=8, GH=4, JH=2. By problem 998, FH=4. Hence, x=√12=2√3.
http://s22.postimg.org/iff0mclox/problem_1000.pngLet P is the midpoint of DEAs the result of problem 998 we haveTriangle DBC congruent to ABE=> AF=DC=8And triangle PFH and FHG are equilateralSince G and H are midpoints of AC and EC => GH= 1/2AE=4In equilateral triangle FGH altitude FJ= GH.sqrt(3)/2= 2.sqrt(3)