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 897.

## Monday, July 8, 2013

### Problem 897: Intersecting Circles, Common External Tangent, Secant, Congruence.

Labels:
common tangent,
congruence,
intersecting circles,
secant

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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