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- Jan 30th 2010, 03:47 PM #1

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- Jan 30th 2010, 04:12 PM #2

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- Jan 30th 2010, 05:18 PM #3
That's false.

$\displaystyle \triangle CDE$ is actually equilateral, thus $\displaystyle \measuredangle\,AEC=75^\circ.$ On the other hand we have that $\displaystyle \triangle BDF$ is also equilateral, then $\displaystyle \measuredangle\,D=90^\circ$ hence $\displaystyle \measuredangle\,FED=45^\circ.$

Finally $\displaystyle \measuredangle\,AEC+\measuredangle\,E+\measuredang le\,FED=180^\circ,$ which implies that $\displaystyle A,\,E$ and $\displaystyle F$ are collinear.

- Jan 30th 2010, 06:03 PM #4

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