Geometry Problem

Click the figure below to see the complete problem 474 about Parallelogram, Diagonal, Circle, Vertex.

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Complete Problem 474

Level: High School, SAT Prep, College geometry

## Saturday, July 17, 2010

### Problem 474: Parallelogram, Diagonal, Circle, Vertex

Labels:
circle,
diagonal,
parallelogram,
vertex

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Let BG intersect AC at I . We have IB=IC and IA=IC

ReplyDeleteLet circumcircles of Triangles DFG and AEG intersect AC at N and M

Connect GE and GF

1. Note that Angle(GFC)=Angle(GND) both angle supplement to angle(NGF)

Angle(CEG)=Angle(BMC)

2. Angle(BMI)=Angle(IND) both angle supplement to angle(GND)

And BM//DN

3. We have Tri. BMI congruence with Tri. NDI ( case ASA) so IM=IN

4. We have CE.CB+CF.CD=CG.CN+CG.CM= CG*(CN+CM)

5. Replace CN+CN=CA we get the result

Peter Tran

vstran@yahoo.com

I'm INDSHAMAT from University of Kelaniya,Srilanka.This is my solution....

ReplyDeleteAngle ACD=Angle CAB(ABCD is a parallelogram)

Angle ACD=Angle GEF and Angle ACB=Angle EFG(GFCE is a cyclic quadrilateral)

Hence GEF and ABC triangles are similar.

So EF/CA=GE/AB=GF/CB,Therefore EF=CA.GF/BC(1) and GE=AB.GF/CB(2)

From Ptolemy's theorem CG.EF=CE.GF+GE.FC

Replacing from (1)&(2)for EF and GE we have

(CG.CA.GF)/CB=CE.GF+(AB.GF.FC)/CB

Excluding GF from both sides, CA.CG=CB.CE+CD.CF

There are some typo errors in my previous comment. Below is my correction.

ReplyDeleteLet BD intersect AC at I . We have IB=ID and IA=IC

Let circumcircles of triangles DFG and BEG intersect AC at N and M respectively

Connect GE and GF

1. Note that Angle(GFC)=Angle(GND) both angle supplement to angle(GFD)

Angle(CEG)=Angle(BMC)

2. Angle(BMI)=Angle(IND) both angle supplement to angle(GND)

And BM//DN

3. We have Tri. BMI congruence with Tri. NDI ( case ASA) so IM=IN

4. We have CE.CB+CF.CD=CG.CN+CG.CM= CG*(CN+CM)

5. Replace CN+CN=CA we get the result