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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 954.
Let AD and BF intersects at P Consider cyclic quad ADBF,G is the pole of PE and E is the pole of PG and hence EG is the polar of PBy result in 953, OH perpendicular to EG so it must be the pole of BF and hence the intersection point tangents B and F.Remaining is straight forwardQED
To William: Why B and F are tangency points?
property of pole and polar ? (H is the pole of BF relative to circle O)
http://img829.imageshack.us/img829/9363/2os8.pngIn cyclic quadrilaterals AHBE and DBAF ( see solution of pro 953) we have ∠ BHE)= ∠ (BAE)= ∠ (FAG)= ∠ (FHG)And OH ⊥GE per problem 953 so HO is angle bisector of ∠ (BHF)
<D=<BAE=<BHE making GHBD cyclic. That makes <G=<HBE=<HAE making GFAH cyclic. It follows that <FHG=<BHE=<FAG=<D. <FOB=2<D and <FHB=180-<D-<D=180-<FOB, yet again making FHBO cyclic. But then <FHO=<BHO because they subtend chords of equal length (radius of O). This combined with <FHG=<BHE says that OH is perpendicular to GE and bisects <FHB.
First we note GH.GE = GA.GB = GF.GD. So DEHF is a cyclic quadrilateral.Thus angles BHE = BAE = FDB = FDE = FHG.Let R be the radius of the circle (O).Now EH^2 - GH^2 = (EH + GH)(EH-GH) = EG.(EH - GH) = EG.EH - EG.GH = EG.EH - GH.GE = EG.EH - GA.GB = Power of E w.r.t Circle (O) - Power of G w.r.t. Circle (O) = (OE^2 - R^2) - (OG^2 - R^2) = (OE^2 - OG^2).Follows that OH is perpendicular to GE.Hence angles OHB = 90 - BHE = 90 - FHG = OHF.