All you need to do is construct a parallelogram!

by v_Enhance, Dec 12, 2011, 6:56 AM

Let $ABC$ be a triangle with cevians $AM$, $BD$ and $CE$ concurring at $P$. If $AM$ is a median, show that $DE \parallel BC$.

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6 Comments

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No Ceva plox :P

by v_Enhance, Dec 12, 2011, 6:56 AM

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This is trivialized with Ceva.

by AwesomeToad, Dec 12, 2011, 6:53 PM

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Ceva

707

by AwesomeToad, Dec 12, 2011, 7:01 PM

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What he intended

by Jinduckey, Dec 13, 2011, 12:19 AM

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Yeah, that was it.

Alternatively

by v_Enhance, Dec 13, 2011, 12:23 AM

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that's what Ms. Nal did at MathPath

by yugrey, Dec 14, 2011, 12:08 AM

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