Proposed Problem
Click the figure below to see the complete problem 394 about Square, 90 Degree Arc, Diagonal, Congruence.
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Complete Problem 394
Level: High School, SAT Prep, College geometry
Tuesday, November 24, 2009
Problem 394: Square, 90 Degree Arc, Diagonal, Congruence
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http://geometri-problemleri.blogspot.com/2009/11/problem-51-ve-cozumu.html
ReplyDeleteEF = a-a/V2 and FC=V2a-a where V=square root. So Another way to prove this w/o any construction would be: a=square side. EF/FC=a(1-1/V2)/a(V2-1)=1/V2 and ED/DC=(a/V2)/a =1/V2. Thus,ED/DC=EF/FC or DG bisects angle BDC. Therefore. BG/GC=DB/DC= aV2/a=V2/1 or BG/(BG+GC)=V2*/(V2+1)or rationalizing, BG =V2*a(V2-1) = 2a-aV2.
ReplyDeleteNow, EF=a-a/V2 or 2EF=2a - aV2. Hence, BG=2*EF.
Ajit
let us find the angles of angleBDG,BGD,DFE,EFD,DGC,GDC. then we can observe that DG is the angular bisector of angle BDC,and tr DEF, tr DGC are simalar triangles using these two concepts we can prove the required
ReplyDeletesuggest to others
ReplyDelete(i did not understand anything from third comment)
a easy way: use EF as middle line for any triangle