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
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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