Geometry Problem

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

Click the figure below to see the complete problem 667.

## Friday, September 9, 2011

### Problem 667: Intersecting Circles, Secant, Tangent, Metric Relations, Mind Map

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http://img716.imageshack.us/img716/570/problem667.png

ReplyDeletePer the result of problem 666, quadrilateral CEDB is cyclic

So (ECD)=(EBD)= β = (EDF)

And (CBF)=(CDE)= α

In triangle EGD we have (CGD)=(GDE)+(GED)

= β+ ( 180- α- β)=180- α

So (CGF) supplement to (CBF) and quadrilateral CGFB is cyclic

We have EF.EB=EG.EC= ED^2

And x^2=4 x 9= 36 ; x=6

Peter Tran

Let EB intersect circle (O) again at H.

ReplyDeleteJoin CB, AB

∠BCG

= ∠BCA + ∠ACG

= ∠BCA + ∠CBA

= ∠BAD

= ∠BFD

= ∠GFE

So G, C, B, F are concyclic

Hence x² = EF.EB = EG.EC = 4 x 9 = 36

∴ x = 6

< ECD = < ABC hence < ECB = < BAD = < BFD implying that BCGF is con cyclic

ReplyDeleteSo EG. EC = EF. EB = ED^ 2 from which

x^2 = 4(4+5) and so x = 6

Sumith Peiris

Moratuwa

Sri Lanka