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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 705.
Happy New Year to all - I've a suggestion for Peter, Pravin and others. We've in our diagram E as the midpoint of GH and similarly F is the midpoint of MN. Thus, AH=AG and AM=AN.What I cannot prove is that AG=AM & in fact AG=AH=AM=AN. In other words, a circle with centre at A and radius = AN will pass thru N,G,M & H and thus /_X=36 since/_NHG=36.Now there remains a small matter of proving AG=AM where your expertise is solicited.CheerioAjit
Continuing the reasoning of Ajit: In right triangle AMD: AM^2 = AF * AD ---(1)In right triangle AHC: AH^2 = AE * AC ---(2) Angle(ADE)=ACE) = 90–58 = 32 degree. Then ΔADE~ΔAFC ---> AF/AE = AC/AD So AF * AD = AE * AC ---(3)By (1),(2)and(3) ---> AM = AH.So AH = AG = AM = AN, therefore the quadrilateral HMGN is cyclic.Then angle (NHG)=(NMG)=36 degrees. So x = 36 degrees.
GX*HX = CX*FX = DX* EX = MX* NXHence H,M,G,N is concyclicHence angle XMG = angle XHN = 36
NOTE : X is the point where EG intersects MF