Prove that the midpoint quadrilateral (a quadrialteral obtained by connecting consecutive midpoints of sides) of an isosceles trapezoid is a rhombus.
I know of three different proofs. Would love to know if there are more.
Update Here is one remaining proof that I know of. Well, sort of proof :) more of an idea.
14 comments:
e--
you might not see this for awhile but thank you for your support; now all i need to know is that e^10000!!
The Dept of Ed has been checking my blog - I can see when (today around 10:55) they visited, for how long (22 minutes), how many page views (10) and where they clicked on exit (compelling comments which linked to their own site!). They are scoping me out to see if I have any credibility and could pose a public relatons threat, which is really all government people ever really care about anyway! As you can tell, I'm 'mad as hell and I'm doing something about it!' I'm certainly not intimidated by anyone from Professors Emeriti to anyone in Washington. I suspect you're similar but most educators are afraid of their own shadows and/or losing their jobs.
i find your observations thought-provoking; i sense you're coming from a perspective that is somewhat different from most of my readers but not unlike mine. How did you find my blog? I see U of Mich and Ann Arbor as locations of someone visiting so I'm guessing that's you, but what is drawing you to my site. It can't be my math problems, which for you must be trivial...
dave
how did i find your blog? it could have been tracing comments from either dan meyer's blog, or i found out from darren kuropatwa. i can't say for sure, i've been to too many places lately. i started paying more attention when i saw stuff about math panel :) if you'd like to know more, we can email.
e
e--
i'd like that; now how do we share email addresses without loss of privacy? (although i haven't exactly been hiding my identity, have i!!)...
btw, how amazing is dan! he has a gifted creative mind and transmits a force field that draws one in; ok that's a bit extreme, but he is unique!
dave
dave, i'm confused. i told you i had cc'd you on the email i had sent to flawn (granted i don't even know if you use that email anymore, and i don't know where i found it, but it was rih one), so you should have my email address. so much for guarding one's identity :)
oops, that was pretty dumb, since i forgot that i had left my email on my profile and you already emailed me! how embarassing!
:)
I hate teaching geometry.
1a. Choose an arbitrary isosceles trapezoid, and throw it on a coordinate plane. (2a,0), (2b,2c), (-2b,2c), (-2a,0). The midpoint quadrilateral: (a+b,c), (0,2c),(-a-b,c),(0,0)
Diagonals have the same midpoint (0,c) --> the thing is a parallelogram.
Diagonals are perpendicular (1 vert, 1 horiz) --> thing is a rhombus.
1b. (could also use the distance formula 4 times: sqr[(a+b)^2 + c^2] pops up each time.
2. No coordinates.
Consider Iso Trap EFGH, with EF = GH and midpoints WXYZ.
Consider diagonal EG. WX || EG (in triangle EFG, WX cuts EF and FG proportionally, must be parallel to the base. Also, YZ || EG (from triangle FGH, same reasoning). Therefore YZ||WX.
Repeat steps on diagonal FH to establish WZ||YX.
We have a parallelogram.
Angle E = Angle H, EZ=EH, EW=HY so ZEW = ZHY and ZW = ZY. By transitivity and properties of a parallelogram, all 4 sides are congruent.
3. Consider an isosceles trapezoid IJKL with IJ = KL. construct midpoints of IJ and KL (M and N) and drop perpendiculars through each to IL. Also extend JK to intersect the perpendiculars.
Consider the 4 new points of intersection. They form a rectangle. M and N are midpoints. Symmetry gives us the same midpoints as for IL and JK. Now the midpoint quadrilateral forms 4 congruent right triangles, and must be a rhombus.
Your set-up forces me to blog with my dead blogger account. Any chance of altering that?
jd2718
Jonathan,
thanks. I never think of proving anything using coordinates. I have one more proof, I'll post solution little later (I want to draw a pic, it's easier).
As for the comments: would it help if I allow anonymous comments? I'm open to other solutions, I'll look into it (right now I don't have an answer, but I'll try).
e
on some blogger sites I am allowed
1. blogger id
2. other
3. anonymous
on others:
1. blogger id
2. other
and here:
1. blogger id
I like the middle option.
Oh, boy, coordinates can be very very useful as a problem solving tool. You have got to try them.
And I am writing about modifying the level of difficulty of problems, (Dave talks about this as well) and will use this one as an example, unless you object.
press here, not up top, to get to me
Jonathan,
I tried changing the permissions, hopefully it'll be easier to comment. I just saw your modifications to the problem, I like them. I have to think about K-5 modification more. Oh, and I know where you are, no need to leave elaborate links on the bottom :) I still owe you the last proof (I get back tomorrow evening, and I'll do it then).
It works!
On K-5, finding areas by graphing on oaktag, cutting and moving? It might take a bunch of tries to make the right cuts.
Or this may be a problem (started as a multi-proof challenge!) that does not translate down to grade school.
Jonathan,
I guess if you use a rectangle instead of a trapezoid it would be much easier for elementary kids to see how all those "corner" triangles are congruent.
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