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Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view more details of problem 1180.
Let AB, DC meet at H so that E is the incentre of Tr. AHD and < BGE = < DGE = 30.Also easily < ACD = 60 + A/2 = < DBG ( A + 60 - A/2) making BECH concylicHence BE = CE = 2FE since Tr . ECF is 30-60-90.Further Tr. s GED is congruent to Tr. CED case ASA making EG = EC
I am not sure from "E is the incentre of Tr. AHD" , you get " < BGE = < DGE = 30.please explain .Peter
Just a typo replace < BGE = < DGE = 30 with < BHE = < DHE = 30
I could not complete the last lineHence EG = 2FE Sumith PeirisMoratuwaSri Lanka
http://s1.postimg.org/kgkafchkf/pro_1180.pngBA meet CD at MSince ∠(A)+ ∠(D)= 120 =>∠(M)=60 Since EA and ED bisect ∠(A) and ∠(D) => E is in center if trị. MAD and ME bisect ∠((M) and ∠( AED)=120MBEC is cyclic and chord BE=chord EC => trị BED is isosceles Triangle EFC is 30-60-90 triangle => EC= 2. EFTriangles ECD and EDG are congruence…. ( case ASA) => EG=EC=2. EF
To Sumith & Peter Why Tr CED congrc to EGD
ASA. ED common. <GDE = <CDE and < GED = < CED = 60 which I'm sure u can figure out why.
OK to SumithFrom congr in tr BGC, GF median and BM median (M on GC)=>EG = 2FE
Interestingly Tr. BGC is equilateral from which the result EG = 2 FE is obvious.