## Sunday, February 4, 2018

### Geometry Problem 1354: Circle, Tangent Lines, Secant, Chord, Center, Parallel Lines

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below. 1. Is AE*AF=AC^2=AD*AO=>OFED=cyclic=> DF=DG.But triangleOFD=triangleOGD => <ODG=<ODF.So OD=perpendicular in FG.Therefore FG//BC.

1. It is not clear to me how a cyclic quadrilateral OFED will give DF=DG in your solution.

Peter

2. is DGF=DFG=>DG=DF (not from quadrlateral OFED)

2. https://photos.app.goo.gl/bNpiPWF2vo2hD4S63
connect OE, OF, DE and DF
since AC and AB tangent to circle O
so OC^2=OE^2=OF^2=OD.OA
or OE/OA=OD/OE and OF/OA=OD/OF
triangle ODE similar to OEA ( case SAS)
and EA/DE=OE/OD => EA=OE/OD x DE… (1)
triangle ODF similar to OFA… ( case SAS)
and FA/DF=OF/OD => FA= OF/OD x DF… (2)
Divide (1) to (2) and note that OF=OE we get
EA/FA=DE/DF => ODA is the internal bisector of angle GDF
So DO ⊥ GF => GF//BC

3. Since OC2 = OD.OA = OE2,
< OED = < OAE
Similarly < OFD = OAE,
So OFED is concyclic.

Hence < GDO = < OFE = < OEF = < ODF = λ say
Therefore < FOE = 180 - 2λ and < FGE = 90 – λ = < GDB

So FG // BC

Sumith Peiris
Moratuwa
Sri Lanka