﻿ Given the right trapezoid ABCD , AB\parallel CD , AB\perp AD , AB=AD=\dfrac{1}{2}CD=1 . Now

#### Question

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 .

Prove AM\parallel plane BEC ;

Prove BC\perp plane BDE ;

Find the distance from D to plane BEC

Collected in the board: Solid Geometry

Steven Zheng posted 3 years ago

Plot MG \parallel EC in the \triangle 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}

Steven Zheng posted 1 year ago

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