∴△ABC and △BCD are isosceles triangles with * * side BC, so point D is at the height of △ABC side BC (also the point of the middle line);

Then sin∠CAD=(BC/2)/AC=m/2.

∠ ACD =∠ Abd = 30。

△ ∴ in ACD comes from sine theorem, AD: SIN ∠ ACD = CD: SIN ∠ CAD.

Therefore, CD = mad (0 < m < 2).

Maybe I won't, but at least I can't solve it with junior high school knowledge.

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