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levietdung1998



\[\begin{array}{l}
\tan \widehat {IDC} = \frac{{IN}}{{DN}} = \frac{3}{2}\\
\tan \widehat {IDC} = \cot \widehat {DCO} = \tan \widehat {ECB}\\
EB = BC.\tan \widehat {ECB} = 4.\frac{3}{2} = 6\\
\to AE = EB - AB = 2\\
E{D^2} = A{E^2} + A{D^2} \to ED = \sqrt {20}
\end{array}\]

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