## Tuesday, July 9, 2013

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

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

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

Click the figure below to see the dynamic geometry demonstration of problem 901. 1. Observe that
(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.

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

1. Sorry i miss a small step in my previous solution,
Extend CD to point X, X is on the other side of C,
∠HDC = ∠JDX = ∠DEJ = ∠DBJ

2. Note to William Fung
Your statement ∠HDC = ∠JDX is true only if H, D, J are collinear.

Peter Tran

3. 3. http://img853.imageshack.us/img853/4403/fso.png

Let 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

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