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