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Quadrilaterals and Diagonals

Date: 18 Jan 1995 13:32:30 -0500
From: Art Mabbott
Subject: Dr. Math

If the diagonals of a quadrilateral are congruent, must the quad be a
rectangle or an isosceles trapezoid?  In both cases, the "sides" that are
not the bases would have to be congruent.

Art Mabbott

Date: 19 Jan 1995 20:39:34 GMT
From: Dr. Math
Subject: Re: Dr. Math

Hey Art,

We are here at Dr. Math, and we are glad to get your question. 
Unfortunately, I think that the answer is no.  I think the easiest way to
see that is not to focus on the fact that the two sides are required to
be congruent but that isoceles trapezoids have two parallel sides.  I
think if you play around a little bit with two crossing pencils you will
quickly see that you can form the tips into a quad that has no parallel
sides.  Hope that visual can be helpful.

Ethan Doctor On Call

Date: 20 Jan 1995 10:59:20 -0500
From: Art Mabbott
Subject: Dr. Math (correction)

Dr. Math:

If one set of sides of a quadrilateral is parallel (the Quad is at 
least a trapezoid), and the diagonals are congruent, must the quad 
be an isosceles trapezoid or a rectangle? 

Jonathan Royalty
(Art Mabbott)

Date: 20 Jan 1995 15:07:20 -0500
From: Dr. Ethan
Subject: Re: Dr. Math (correction)


Well now we are cooking!  Isn't it neat how if you add one 
little requirement, we get the result?

The answer is YES.  If one pair of opposite sides of a quadrilateral 
are parallel and the diagonals are congruent, then the quadrilateral 
is an isoceles trapezoid (You don't have to say rectangle because a 
rectangle is an isosceles trapezoid).

Here is the proof:

     A________________B      Consider the quad labeled as shown.
    /                  \     Then AB is parallel to DC and
   /                    \    AD is congruent to BC.
  /                      \

So now consider the Triangle DBC and the Triangle CAD.  We know that AC 
is congruent to DB and that DC is congruent to DC.  If we could just show 
that Angle ACD and Angle BDC were congruent then we could show that 
the triangles in question were congruent by the side angle side postulate.

        To do this, drop two perpendicular lines from A and B (call their
points of intersection E and F respectively).

     A_______________B       Now examine the triangles CAE and DBF.
    /|               |\      AE is congruent to BF and  
   / |               | \     AC is still congruent to BD.
  /  |               |  \    Because it is a right triangle,
D/___|_______________|___\C  Triangles CAE and DBF are congruent
         E               F

Almost there!

This gives us that Angles ACD and BDC are congruent, and 
that was what we needed to show that triangles ACD and BDC were congruent. 

Since those triangles are congruent, AD and BC are congruent.  That was 
what we wanted to prove.  If you will go back to the beginning, you will
notice that all we assumed was one parallel pair and congruent diagonals.

Hope that helps.
Ethan, Doctor On Call

Date: 20 Jan 1995 15:29:16 -0500
From: Art Mabbott
Subject: Re: Dr. Math (correction)

Dr. Ethan:

After all the discussion on the Geometry Forum and the NCTM-L
discussion groups, I'm surprised that you would tell poor Jonathan that a
rectangle is an Isosceles Trapezoid.  There is not agreement on that and
in fact there are camps out here that still define parallelograms and
trapezoids as disjoint subset of the Quads. 

Art Mabbott
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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