## Friday, January 3, 2014

### Geometry Problem 954: Intersecting Circles, Secant, Common Chord, Cyclic Quadrilateral, Angle Bisector

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

Click the figure below to see the complete problem 954.

1. Let AD and BF intersects at P
G is the pole of PE and E is the pole of PG and hence EG is the polar of P
By 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 forward
QED

1. To William: Why B and F are tangency points?

2. property of pole and polar ?
(H is the pole of BF relative to circle O)

2. http://img829.imageshack.us/img829/9363/2os8.png
In 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)

3. <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.

4. 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.