Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 705.
Sunday, December 18, 2011
Problem 705: Intersecting Circles, Diameter, Angles, Measurement
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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.
ReplyDeleteWhat 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.
Cheerio
Ajit
Continuing the reasoning of Ajit:
ReplyDeleteIn 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.
Thanks Anonymous.
ReplyDeleteGX*HX = CX*FX = DX* EX = MX* NX
ReplyDeleteHence H,M,G,N is concyclic
Hence angle XMG = angle XHN = 36
NOTE : X is the point where EG intersects MF
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