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

Labels:
angle,
diameter,
intersecting circles,
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

Delete