TITLE: Determining whether the shape is a rectangle
QUESTION [0 upvotes]: While solving a problem, I came across a little hump which is impeding a pure solution. If there is a quadrilateral ABCD where $\angle B = 90^\circ$ and $AD = BC$ and $\angle D = \angle C$, is it possible for the shape to be anything other than a rectangle? I don't believe it can be anything else, but I need to be absolutely certain as my proof for another problem depends on this result.

REPLY [0 votes]: Triangles DCB, CDA are congruent (SAS).So ADCB is an isosceles trapezium,making  AD = BC.  Drop perpendiculars AD2 BC2 as perpendicular heights onto DC.
The perpendiculars are equal, due to congruence of triangles AD2D and AC2C. 
So now AB is parallel to CD. B + C = 180 degrees, by virtue of Euclid parallel lines /transversal cutting theorem. Since B is given 90 deg, C is 90 degrees.
So also D and finally the remaining corner angle A is 90 degrees.
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