Quadrilaterals and DiagonalsDate: 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) Ah, 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. / \ D/________________________\C 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 |
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