Given the tetrahedron ABCD, O , E are the midpoints of BD and BC , respectively. CA=CB=CD=BD=2 , AB=AD=\sqrt{2} .

Prove AO\perp plane BCD ;

Find the cosine of the angle formed by the straight lines AB and CD in different planes.

Find the distance from E to plane ACD

Collected in the board: Solid Geometry

Steven Zheng posted 2 years ago


\because AB=AD=\sqrt{2} and point O is the midpoint of BD

\therefore \triangle ABD is a an isosceles triangle and

AO\perp BD

Using Pythagorean Theorem

AO = \sqrt{AB^2-(\dfrac{BD}{2})^2 } =\sqrt{2-1}=1

On the other hand,

\because CB=CD=BD=2

\therefore \triangle BCD is an equilateral triangle and

OC\perp BD

Using Pythagorean Theorem

OC = \sqrt{BC^2-(\dfrac{BD}{2} )^2} =\sqrt{4-1}=\sqrt{3}

For three sides of \triangle AOC ,

CA=2, OC=\sqrt{3}, AO=1,


Therefore, \triangle AOC is a right triangle

AO\perp OC

Since AO is perpendicular to two intersecting lines in the plane of BCD,

AO \perp plane BCD

Draw a segment BF from point B which is in parallel with DC in the plane BCD and BF = CD=2

\because cd\parallel BF

\angle CBF = \angle BCD = 60\degree

\angle OBF = \angle CBF +\angle CBD = 120\degree

Apply the law of cosines for \triangle OBF

OF^2=BO^2+BF^2-2BD\cdotp BF\cdotp \cos \angle OBF

=2^2+1^2-2\cdotp 2\cdotp 1\cos 120\degree


On the other hand,

\because AO\perp plane BCD

\triangle AOF is a right triangle

AF^2 = AO^2+OF^2 =8

Apply the law of cosines for \triangle ABF

\cos\angle ABF =\dfrac{ BF^2+AB^2-AF^2}{2AB\cdotp BF }=\dfrac{2^2+(\sqrt{2})^2-8}{2\times 2\times \sqrt{2} } =-\dfrac{2}{4\sqrt{2} }

=-\dfrac{\sqrt{2} }{4}

Steven Zheng posted 1 year ago

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