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

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

Uniteasy Admin · Accepted Answer

$$\because AB=AD=\sqrt{2}   $$ and point $$O$$ is the midpoint of $$BD$$
$$	herefore   	riangle ABD$$ is a an isosceles triangle  and
$$AO\perp BD $$n
Using Pythagorean Theorem 
$$AO = \sqrt{AB^2-(\dfrac{BD}{2})^2  } =\sqrt{2-1}=1   $$
On the other hand,  
$$\because   CB=CD=BD=2 $$  
$$	herefore   	riangle  BCD$$ 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 $$	riangle AOC $$,
$$CA=2, OC=\sqrt{3}, AO=1,$$
$$CA^2=OC^2+AO^2$$
Therefore, $$	riangle AOC $$ is a right triangle
$$AO\perp OC $$n
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 $$	riangle OBF $$
$$OF^2=BO^2+BF^2-2BD\cdotp BF\cdotp \cos \angle OBF $$
$$=2^2+1^2-2\cdotp 2\cdotp 1\cos 120\degree    $$
$$=7$$
On the other hand,
$$\because AO\perp$$ plane $$BCD$$
$$	riangle AOF$$ is a right triangle 
$$AF^2 = AO^2+OF^2 =8$$
  Apply the law of cosines for $$	riangle ABF $$
$$\cos\angle ABF =\dfrac{ BF^2+AB^2-AF^2}{2AB\cdotp BF }=\dfrac{2^2+(\sqrt{2})^2-8}{2	imes 2	imes \sqrt{2}    }   =-\dfrac{2}{4\sqrt{2}  }   $$
$$=-\dfrac{\sqrt{2}  }{4}  $$

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