Question 1063775
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I will just give an outline of how to prove it. You
can write it up in a two-column proof.

A rhombus is a parallelogram with four congruent sides. 
So, all sides of rhombus ABCD are congruent. That is 

AB &#8773; BC &#8773; CD &#8773; AD 

   {{{drawing(150,200,-3,3,-4,4,line(-2,0,0,3),line(0,3,2,0),

line(-2,0,0,-3),line(0,-3,2,0),line(0,-3,0,3),line(-2,0,2,0),
locate(-2.35,0.3,B),locate(2.1,0.3,D),locate(-.1,3.5,A),locate(-.1,-3,C),
locate(.1,0,E)
 )}}}
 
We also know that the diagonals of a parallelogram bisect 
each other. Since a rhombus is a parallelogram, it also has
this property. 

Therefore BE &#8773; DE, AE &#8773; CE.

Therefore all 4 triangles are congruent by SSS.

Edwin</pre></b></font>