See complete Problem 190 at:
www.gogeometry.com/problem/p190_tangent_circle_diameter_perpendicular.htm
Tangent circles, Tangent chord, Perpendicular, Distance. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Monday, October 13, 2008
Elearn Geometry Problem 190, Circle
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DB/2r=EO'/AO'=1/3, x/DB=1/3, so x=2r/9
ReplyDeleteangles FDB=DAB, (pependicular sides )=> right triangles AEO'and DFB are similar
ReplyDelete=> x/r=DB/3r
( DB=4/3r because triangles AEO' and ADB are similar, EO'//DB)
=>x/r=4r/3/3r => x=4r/9 or 2r/9 (r=1/2 R)
connects point D and B.
ReplyDeleteBD^2 = FB . AB
(2r/3)^2 = x . 2r
4r^2/9 = x . 2r
then x = 2r/9
Let r=1 for simplicity
ReplyDeletem(O'ED) and m(O'FD) are supplimentary angles => O'EFD are concyclic and m(EO'A)=m(EDF)
hence triangles AEO',AFD and ADB are similar right triangles
AE^2=AO*AB (Since AE is tangent to O')
=> AE=Sqrt(2)
Since AEO' and ADB are similar, we have
AD=4√2/3
Similarly AEO' and AFD are similar,
AF=AE*AD/AO'
=> AF=(√2)*(4√2/3)/(3/2)
=> AF=16/9
Hence BF=AB-AF=2-16/9=2/9
Since O'E = r/2, O'E = O'A/3
ReplyDeleteBut O'E // BD, so BD = AB/3 = 2r/3
Now BD^2 = 2r.x i.e. 4r^2/9 = 2r/3 from which we get
x = 2r/9
Sumith Peiris
Moratuwa
Sri Lanka