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 900.

## Tuesday, July 9, 2013

### Problem 900: Intersecting Circles, Common External Tangent, Secant, Angle, Congruence

Labels:
angle,
congruence,
dynamic geometry,
GeoGebra,
HTML5,
intersecting circles,
ipad

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By problem 899, F,B,D,G concylic,

ReplyDelete∠GBD=∠GFD=∠CFA=∠CBA

Quadri. FCAB is cyclic => ∠ (CBA)= ∠(CFA)

ReplyDeleteQuadr. FGDB is cyclic per result of problem 899=> ∠(DBG)= ∠(GFD)= ∠(CFA)

So ∠(ABC)= ∠(DBG)

Since ABFC concyclic, and also BDGF (Problem 899).

ReplyDeleteThus

∠ABC = ∠AFC = ∠DFG = ∠DBG.

Problem 900

ReplyDeleteIn the convex quadrilateral GCADFE by Miquel’s theorem the point B is point Miquel.So the F,G,D and B are concyclic.So <CBA=<CFA=<GFD=<GBD.

MANOLOUDIS APOSTOLIS 4 HIGH SCHOOL OF KORYDALLOS PIRAEUS-GREECE