Given the right trapezoid $$ ABCD $$ , $$ AB\parallel CD $$ , $$ AB\perp AD $$ , $$ AB=AD=\dfrac{1}{2}CD=1 $$ . Now make a square $$ ADEF $$ with $$ AD $$ as one of its sides, and then fold the square along $$ AD $$ to make the plane $$ ADEF $$ is perpendicular to plane $$ ABCD $$ , $$ M $$ is the midpoint of $$ ED $$ .

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Given the right trapezoid $$ ABCD $$ , $$ AB\parallel CD $$ , $$ AB\perp AD $$ , $$ AB=AD=\dfrac{1}{2}CD=1 $$ .  Now make a square $$ ADEF $$ with $$ AD $$ as one of its sides, and then fold the square along $$ AD $$ to make the plane $$ ADEF $$ is perpendicular to plane $$ ABCD $$ ,  $$ M $$ is the midpoint of $$ ED $$ .

Uniteasy Admin · Accepted Answer

Plot $$MG \parallel EC$$ in the $$	riangle DCE$$
Since point M is the midpoint of DE,  point G is the midpoint of CD.
Connect AG. In the right trapezoid ABCD, $$ AB\parallel CD $$ , $$ AB\perp AD $$ , $$ AB=AD=\dfrac{1}{2}CD=1 $$, therefore $$AG \parallel BC$$
Now there are two intersection lines MG and AG in parallel with plane BEC. Then Plane AMG which contains the two lines is parallel with plane BEC.
Hence, $$AM \parallel$$ plane BEC In the right trapezoid ABCD, the triangle BDC is a right isosceles triangle. So $$BC \perp BD$$. On other hand, the square $$ ADEF \perp ABCD$$ as given condition. The $$ DE\perp  ABCD$$,   $$DE\perp BC$$
Now BC is perpendicular to two intersection lines on the plane of BDE, hence, it's perpendicular to the plane. Drop an altitude from point D to BE, $$DH \perp BE$$. Since BE\perp plane BDE, $$BE\perp DH$$. Therefore, DH is the distance from D to plane BCE.
In the right triangle BDE,
$$BD=\sqrt{2}, DE = 1 $$, then, $$BE = \sqrt{3} $$
$$DH = \dfrac{BD\cdot DE}{BE} = \dfrac{\sqrt{2} }{\sqrt{3} } = \dfrac{\sqrt{6} }{3} $$

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