Geometry Problem

Click the figure below to see the complete problem 527 about Circle, Tangent, Secant, Chord, Midpoint, Congruence, Measurement.

See also:

Complete Problem 527

Level: High School, SAT Prep, College geometry

## Sunday, October 10, 2010

### Problem 527: Circle, Tangent, Secant, Chord, Midpoint, Measurement.

Subscribe to:
Post Comments (Atom)

See below link for details http://img109.imageshack.us/img109/442/problem527.png

ReplyDeleteLet N is the intersection of GD and CE.

1. We have GN.DN=NC.NE and HG.HD=FE.FC=HB^2=FB^2 (power of the point theorem)

Consider triangle NHF and 2 secants GEM and ACD . Apply Menelaus’ theorem twice:

(MF/MH).(GH/GN).(EN/EF)= 1 and (AF/AH).(DH/DN).(CN/CF)=1

2. Multiply each side of above equations we get

(MF/MH).(AF/AH) .(GH.DH)/ (GN.DN) .(EN.CN)/(EF.CF)=1 (1)

But GN.DN=EN.CN and GH.DH=EF.CF ( from step 1)

equation (1) become (MF/MH).(AD/AH)=1 (2)

3 Let BH=BF=b and BA=x , BM=y

Equation (2) become (y-b)/(y+b) * (x+b)/(x-b)= 1

Develop and simplify we will get 2.b.y=2.b.x

So x=y and B is the midpoint of AM

Peter Tran