Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Details: Click on the figure below.
HC=2, => BH=6 => AH²=100, HK²+AK²=100 => HK=5√2
Applying the knowledge from the previous problems, we haveAH=HM=GC=10 ----------(1)and AHMG is cyclic with m(AHM) = 90-----------(2)=> AM=10Sqrt(2) ---------(3)=> K is the center of the circle on which A,H,M,G lie=> HK=AM/2 = 5sQRT(2)
Problem 1331Suppose that the BC intersects GM in point N then NM=FG=6, or MG=8-6=2.AM^2=AG^2+MG^2=14^2+2^2=200, or AM=10√2 .So HK=AM/2=5√(2 ).
AB = 8, BH = 6 so AH = 10Tr. HAK is right isosceles Hence HK = 10/sqrt2 = 5sqrt2Sumith PeirisMoratuwaSri Lanka