Geometry Problem

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

Click the figure below to see the complete problem 953.

## Friday, January 3, 2014

### Geometry Problem 953: Intersecting Circles, Secant, Common Chord, Cyclic Quadrilateral, 90 Degrees, Perpendicular

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ReplyDeleteIn cyclic quadrilaterals AHBE and DBAF we have ∠ BHE)= ∠ (BAE)= ∠ (FDB)

since ∠ (BHE)= ∠ (GDB) => quadrilateral DBHG is cyclic and EB.ED=EA.EF=EH.EG

So quadrilateral GFAH is cyclic.

Let R =radius of circle O

Power of G to circle O= GO^2-R^2= GA.GB=GH.GE (1)

Power of E to circle O= EO^2-R^2=EA.EF=EH.EG……. (2)

(1)- (2)= GO^2-EO^2= GE.(GH-EH)= linear function of position of H on GE….. (3)

Let H’ is the perpendicular projection of O over GE

We also have GO^2-EO^2= GE.(GH’-EH’) ….. (4)

Compare (3) and (4) we get H coincide to H’ => OH ⊥GE

There is an easier proof in my opinion.

ReplyDeleteFirst we prove that FGAH is cyclic. Then we prove that FOBH** is cyclic. After that it's pretty simple. As follows

Let ∠ (GDE)= x, so ∠ (FHG)= ∠ (BHE)= x; ∠ (FOB)=2x (angle at the centre)

∠ (OFB)= 90-x (base angle of isosceles. Hence ∠ (OBH)= 90-x (angles on same segment)

finally, ∠ (OHE)=∠ (BHE)+ ∠ (OBH)= 90-x+x =90, as required

**NB.

Let ∠ (GDE)= x, so ∠ (FOB)=2x

∠ (FAG)= x, so ∠ (FHG)=x

∠ (BAE)= x, so ∠ (BHE)=x

Now, 2x+∠ (FHB)= 180, but ∠ (FOB)=2x

so ∠ (FOB)+∠ (FHB)=180, hence FOBH is cyclic.