Geometry Problem. GeoGebra, HTML5 Animation for iPad and Nexus.
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Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the dynamic geometry demonstration of problem 897.
Monday, July 8, 2013
Problem 897: Intersecting Circles, Common External Tangent, Secant, Congruence.
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Let Dx and Cy are the extension of CD and DC
ReplyDeleteWe have ∠(DAE)=∠ (EDx)= ∠(CDG)
and ∠(FAC)= ∠(AFy)= ∠(DGC)
But ∠(DAE)= ∠(FAC) => ∠(CDG)= ∠(DGC)
So GDC is isosceles triangle and GC=GD
∠DCA=∠CFA
ReplyDelete∠CFA+∠CAF+∠ACF=180deg
∠GCD+∠DCA+∠ACF=180deg ---> ∠GCD=∠CAF
similarly, ∠GDC=∠DAE
∠CAF=∠DAE, so ∠GCD=∠GDC ---> GC=GD
AngleGCD = CDA + CFA = DEA + DCA = GDC
ReplyDeleteSo Triangle GCD isosceles