Geometry Problem. GeoGebra, HTML5 Animation for iPad and Nexus.

Post your solution in the comments box below.

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

Click the figure below to see the dynamic geometry demonstration of problem 901.

## Tuesday, July 9, 2013

### Problem 901: Intersecting Circles, Common External Tangent, Secant, Circumcircle, Concyclic Points, Cyclic Quadrilateral

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Observe that

ReplyDelete(1) ∠HAD = ∠DJB

(2) ∠HDA = ∠CDA + ∠HDC = ∠AED + ∠DBJ = ∠ABJ

So ∆HAD~∆HJB, with power of cirlce, H-D-J collinear.

Hence,

∠HJB = ∠DJB = ∠FAB = ∠FCB

by the converse of ext.∠ of cylic quad, Q.E.D.

Can you explain why ∠HDC = ∠DBJ? Thanks.

ReplyDeleteSorry i miss a small step in my previous solution,

DeleteExtend CD to point X, X is on the other side of C,

∠HDC = ∠JDX = ∠DEJ = ∠DBJ

Note to William Fung

DeleteYour statement ∠HDC = ∠JDX is true only if H, D, J are collinear.

Peter Tran

Thanks for your correction

Deletehttp://img853.imageshack.us/img853/4403/fso.png

ReplyDeleteLet HD cut circle O1 at J’

Since H is on radical line AB of circles O and O1

So We have HC.HF=HA.HB=HD.HJ’ => F,C,D,J’ is cocyclic

∠(CFJ’) supplement to ∠(CDJ’)

∠(CDJ’) supplement to ∠(HDC) and ∠(HDC)= ∠(DEJ’)

So ∠(DEJ’)= ∠(CFJ) => G,F,E,J’ is cocyclic => J’ coincide to J

The rest of my solution is similar to W. Fung solution

It seems that the past few problems could give enough hints on how to solve #738. Any thoughts?

ReplyDelete