## Monday, October 13, 2008

### Elearn Geometry Problem 190, Circle

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

1. DB/2r=EO'/AO'=1/3, x/DB=1/3, so x=2r/9

2. angles FDB=DAB, (pependicular sides )=> right triangles AEO'and DFB are similar
=> 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)

3. connects point D and B.
BD^2 = FB . AB
(2r/3)^2 = x . 2r
4r^2/9 = x . 2r
then x = 2r/9

4. Let r=1 for simplicity
m(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

Similarly AEO' and AFD are similar,
=> AF=(√2)*(4√2/3)/(3/2)
=> AF=16/9

Hence BF=AB-AF=2-16/9=2/9

5. Since O'E = r/2, O'E = O'A/3
But 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