There are two D's in the problem, so let the D on AB be \(N\). Let \(O=DM\cap CN\).

We must find
\[[DNO]=[DNC]-[DOC]\]
Since the sidelength is 1, \([DNC]=\dfrac{1}{2}\). Since \(\triangle DNC\sim\triangle DCM\),
\[\dfrac{[DNC]}{[DCM]}=\dfrac{DC^2}{DM^2}=\dfrac{1}{\dfrac{5}{4}}=\dfrac{4}{5}\]
Also, \([DCM]=\dfrac{1}{4}\), so \([DOC]=\dfrac{1}{5}\). Therefore, the answer is
\[\dfrac{1}{2}-\dfrac{1}{5}=\boxed{\dfrac{3}{10}}\]

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TopNewestThere are two D's in the problem, so let the D on AB be \(N\). Let \(O=DM\cap CN\).

We must find \[[DNO]=[DNC]-[DOC]\] Since the sidelength is 1, \([DNC]=\dfrac{1}{2}\). Since \(\triangle DNC\sim\triangle DCM\), \[\dfrac{[DNC]}{[DCM]}=\dfrac{DC^2}{DM^2}=\dfrac{1}{\dfrac{5}{4}}=\dfrac{4}{5}\] Also, \([DCM]=\dfrac{1}{4}\), so \([DOC]=\dfrac{1}{5}\). Therefore, the answer is \[\dfrac{1}{2}-\dfrac{1}{5}=\boxed{\dfrac{3}{10}}\]

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There is something special about triangle U. See if anyone else can figure it out. Hint: right triangle

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.3

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