1966 AHSME Problems/Problem 31

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

1966 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 30	Followed by Problem 32
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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1966 AHSME Problems/Problem 31

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