Just getting started with this proof

[Scorpino]

I don't know where to start on this. If I could just get a few initial steps other than the given, that'd be great.

Given: ST = UV; W, X, Y and Z are midpoints
Prove: WZXY is a rhombus

It stinks because this isn't even an isosceles trapazoid, so I can prove that the top and bottom sides are parallel. Check out the sketchpad picture of it in the attachment.

 

[coomast]

It can be solved in considering all the points in the XY plane as vectors. This means that you choose an origin somewhere in the plane of the figure, inside or outside the STUV figure does not matter. After this draw a classical X and Y axes through this origin point. Every point of the figure has now some coordinates in this axes-system, and can be regarded as the endpoint of a position vector starting in the origin and ending at that point.

Now we can use basic algebra of vectors in showing that the lengths of the "rhombus" are equal. Consider p.e. the position of the point W. This is (omitting the arrow above all symbols for simplicity):

$$W=\frac{1}{2}(S+U)$$

The position of point Y is:

$$Y=\frac{1}{2}(S+V)$$

The length of the line YW is nothing more than the absolute value of the difference of the points W and Y and can be calcuated by considering the vector between Y and W. This vector is:

$$YW}=\frac{1}{2}(S+V-S-U)=\frac{1}{2}(V-U)$$

The length is the absolute value of this vector. The same goes for the other ones, X and Z. Write down the lengths WZ, XZ and XY and use the fact that ST = UV, in showing that the lengths are equal.

 

[uart]

I don't know where to start on this. If I could just get a few initial steps other than the given, that'd be great.

Given: ST = UV; W, X, Y and Z are midpoints
Prove: WZXY is a rhombus

You can't prove that it's a rhombus because it's not (in general). If all you are given is that its a quadralateral with equal length diagonals then sorry but that's not enough and it's easy to come up with counter-examples.

The original problem might have also specified that SU is parallel to TV (something you didn't tell us), in which case you do get a rhombus when connecting those mid-points.

 

[uart]

BTW. coomast's method is still good, you can use it to easily prove that the opposite sides of WZXY are equal length, so it is a parallelogram, but not in general a rhombus.

 

[coomast]

BTW. coomast's method is still good, you can use it to easily prove that the opposite sides of WZXY are equal length, so it is a parallelogram, but not in general a rhombus.

You're right, it is a parallellogram, but not in general a rhombus. I made the mistake by looking only at the following quote (from http://en.wikipedia.org/wiki/Rhombus):

"In geometry, a rhombus (plural rhombi) or rhomb (plural rhombs) is an equilateral quadrilateral. In other words, it is a four-sided polygon in which every side has the same length."

So in order to prove that it is a rhombus one needs to show additionally that the lines WX and YZ are perpendicular. I believe that this is sufficient.

 

[disregardthat]

Isn't it easier to:
Let UV=2a
ST=2b
observe that triangle SYW is similar to SUV, and SW is half the length of SU. Then WY is half the length of UV, i.e WY=a
Use the similar technique on the other sides, and you got yourself a parallellogram.

I seroiously doubt it is a rhombus. If you draw the convex quadralateral with one side much bigger that the opposite, you can see that sides of the parallellogram is much larger than the other.

 

[coomast]

Isn't it easier to:
Let UV=2a
ST=2b
observe that triangle SYW is similar to SUV, and SW is half the length of SU. Then WY is half the length of UV, i.e WY=a
Use the similar technique on the other sides, and you got yourself a parallellogram.

I seroiously doubt it is a rhombus. If you draw the convex quadralateral with one side much bigger that the opposite, you can see that sides of the parallellogram is much larger than the other.

This is indeed possible, there are several methods in showing it is a parallelogram. However in the original post it was given that UV=ST from which a=b and thus all lengths are equal. I looked further on the definition of a rhombus and it seems that it was sufficient to show that all lengths are equal in order to obtain a rhombus. See, p.e.:

http://www.elsy.at/kurse/index.php?kurs=Parallelogram+and+Rhombus&status=public

So my original method was enough to solve the question, the perpendicularity of the diagonals follows from the equal lengths property and vice verse. The use of similar triangles is also valid.

@Jarle: In case this is not correct, can you present a picture where you think it is going wrong?

 

[disregardthat]

Must've overlooked the fact that ST=UV

It's definitely a rhombus.

I believe a rhombus is defined by that it is a quadrilateral where all four sides are of equal length. If not, it must be a property preserved rhombuses.

 

[coomast]

I can't see the picture yet, it's approval is pending, however in following the explanation you gave, I made a sketch and it seems that the lengths ST and UV are not equal. This was given in the original post. Does the counter example has this property?

Maybe I should wait a bit until I can see the picture.

 

[disregardthat]

Now it's easy to construct the two equal length diagonals SV and TU such that points T and V both lie to the right of the locus, as shown in the diagram.

I believe that the question stated that ST = UV and not that SV=TU. These statements are not equivalent.

 

[uart]

Oh yes I misread the question. I thought it was the two diagonals that were supposed to be the same length! I've been looking at the wrong problem the whole time.

I hope nobody minds if I delete my previous reply as it is usless.

 

[coomast]

OK, I consider the problem as solved then. It was a nice "constructive" discussion. :-)