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

Labels:
chord,
circle,
diameter,
distance,
perpendicular,
tangency point,
tangent

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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