Difference between revisions of "1966 AHSME Problems/Problem 31"

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== Solution ==
 
== Solution ==
 
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<math>\fbox{D}</math>
  
 
== See also ==
 
== See also ==

Revision as of 02:33, 15 September 2014

Problem

Triangle $ABC$ is inscribed in a circle with center $O'$. A circle with center $O$ is inscribed in triangle $ABC$. $AO$ is drawn, and extended to intersect the larger circle in $D$. Then we must have:

$\text{(A) } CD=BD=O'D \quad \text{(B) } AO=CO=OD \quad \text{(C) } CD=CO=BD \\ \text{(D) } CD=OD=BD \quad \text{(E) } O'B=O'C=OD$

Solution

$\fbox{D}$

See also

1966 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 30
Followed by
Problem 32
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