Proving a quadrilateral to be a rhombus

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darkmasterz8

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May 13, 2009
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Well I have this homework problem.I drew it out on paint to make it clear.

2mhu2xs.jpg



Ok I know that the parallel lines make it a parallelogram.It has both pairs of opposite sides that are parallel.I think this proves that its a rhombus because a rhombus has all the properties of a parallelogram.My problem is that there is one more given that is useless?Im not sure if you can just leave it there.Maybe Im mistaken or something?

It would be helpful if someone explain what Im doing wrong. :?
 
The rhombus is a quadrilateral with all sides equal in length. So, you have to prove that BG=GE=ED=DB. Can you show that the two triangles are isosceles triangles?
 
darkmasterz8 said:
Isosceles triangle has two congruent sides which I dont know how to find.
Click to expand...

More relevant in this problem is that: In a triangle, if two angles are congruent then it is an isoscales triangle.

Use the "interior angle theorems of parallel lines" to find the congruent angles - hence the isoscales triangles.
 
Loren said:
The rhombus is a quadrilateral with all sides equal in length. So, you have to prove that BG=GE=ED=DB. Can you show that the two triangles are isosceles triangles?
Click to expand...
Did you understand this statement?
 
Ok, so I have to prove all sides are congruent. I just dont know how to use the angle bisector to help me.

Also I find the Rays and Line segments to be confusing as well.

EDIT:I just figured out something.Can BF be the diagonal that bisects their angles which is <DBG and <DEG.Therefore <BDE is congruent to <GBE and <DEB is congruent to <GEB.You get the two congruent angles that you mention to be needed for an isosceles triangle.
 
darkmasterz8 said:
Ok, so I have to prove all sides are congruent. I just dont know how to use the angle bisector to help me.

Also I find the Rays and Line segments to be confusing as well.

EDIT:I just figured out something.Can BF be the diagonal that bisects their angles which is <DBG and <DEG.Therefore <BDE is congruent to <GBE and <DEB is congruent to <GEB.You get the two congruent angles that you mention to be needed for an isosceles triangle.
Click to expand...

What does an angle bisector DO to an angle??

And, if you have parallel lines intersected by a transversal, what do you know about the angles formed (I suggest that you look at the alternate interior angles).

This line of thought will lead you to an important conclusion about the angles in each of the two triangles.

AND, you know you've got a parallelogram...what properties of a parallelogram might be helpful here?
 
darkmasterz8 said:
Well I have this homework problem.I drew it out on paint to make it clear.

2mhu2xs.jpg



Ok I know that the parallel lines make it a parallelogram.It has both pairs of opposite sides that are parallel.I think this proves that its a rhombus because a rhombus has all the properties of a parallelogram.My problem is that there is one more given that is useless?Im not sure if you can just leave it there.Maybe Im mistaken or something?

It would be helpful if someone explain what Im doing wrong. :?
Click to expand...
You'll need to consider properties of two sets of pairs of triangles, namely triangles (BDE & EGB) then triangles (GBD & DEG)

Once you prove all these triangles are isoscales - you are home free.
 
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