Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Sunday, February 4, 2018

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

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Is AE*AF=AC^2=AD*AO=>OFED=cyclic=> DF=DG.But triangleOFD=triangleOGD => <ODG=<ODF.So OD=perpendicular in FG.Therefore FG//BC.

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

DeletePlease explain

Peter

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

Deletehttps://photos.app.goo.gl/bNpiPWF2vo2hD4S63

ReplyDeleteconnect 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

Since OC2 = OD.OA = OE2,

ReplyDelete< 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