Determining whether the shape is a rectangle

Question

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.

Answer 1

It has to be a rectangle, if you say that two sides are equal (AD=BC) and $\angle D= \angle C$, and $\angle B=90^\circ$, This implies that AB=CD, (can be proved by taking origin at B and marking respective coordinates) .

Then you draw diagonal BD, and see that triangles BAD and BCD become congruent (by SSS),

$\implies$ $\angle C=\angle A$, as $\angle C=\angle D$, every angle becomes $90^\circ$(sum is $360^\circ$)

Answer 2

Consider the figure below.

Let assume that $$ \hat{C} + \hat{D} < 180^0 $$ Then draw the line $CB$ which intersects with line $DA$ at $E$. Then we will have $\hat{B_1}=90^0$ which implies that $EA > EB$. This in turn implies that $ED=EA + a > EC = EB + a$. This however is a contradiction, because the triangle $ECD$ should be isosceles.

Symmetric arguments should provide contradiction for the case $\hat{C} + \hat{D} < 180^0$. Thus we conclude $\hat{C} + \hat{D} = 180^0$ and consequently that all angles are right angles.

Answer 3

Ignoring the right angle, we readily deduce that $\square ABCD$ is at least an isosceles trapezoid with bases $\overline{AB}$ and $\overline{CD}$. Recalling that one of the angles is right, we can use some combination of parallelism, symmetry, and/or transitivity arguments to show that the remaining three angles are right, as well.

Answer 4

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.