Prove this is a rectangle Suppose we have quadrilateral $ABCD$ where $m\angle A = m\angle B = 90^\circ$ and $AB \cong CD$. Is this figure always a rectangle? If not, can someone give a counterexample?
I tried drawing the diagonal $BD$ and tried to show $\triangle BDC \cong \triangle DBA$. We are already given $\overline{AB} \cong \overline{DC}$, and the triangles share side $\overline{DB}$.
We know that $AD \parallel BC$ because $m\angle A + m\angle ABC = 180^\circ$. This means $\angle DBC \cong \angle BDA$. But, unfortunately, we still don't know $\triangle BDC \cong \triangle DBA$ because the angle isn't between the two sides that we know are equal.

Even though I can't prove it, I still can't find a counterexample.
 A: Let $D'$ be the point such that $ABCD'$ is a rectangle. Then $D'$ is on $AD$ because $\angle A$ is right. 
From $CD'=AB$ (hold in rectangle) and $AB=CD$ (given), we see that $CDD'$ is an isosceles triangle. Its angle at $D$ is right (it is the angle between $CD$ and the line $AC$), hence the angle at $D$ is also right. Thus three out of four angles in $ABCD$ are right, hence they are all right angles
A: We know that $\overline{AD}$ and $\overline{BC}$ are parallel.
Suppose $E$ is a point on the extended line $\stackrel{\leftrightarrow}{AD}$
such that the length $CE = CD$.
Then $\triangle DCE$ is an isoceles triangle with base $\overline{DE}$.
let $M$ be the midpoint of $\overline{DE}$.
Then $\overline{CM} \perp \;\stackrel{\leftrightarrow}{AD}$
and $CM \leq CD$, with equality only if $D = E = M$.
But now $ABCM$ is a rectangle (having right angles at at least three vertices),
hence $CM = AB$. We are given $AB = CD$, therefore $CM=CD$,
therefore $D = E$, therefore $ABCD$ is a rectangle.
A: Find the point E such that $\triangle ACE$ is congruent to $\triangle ABC$.

Since $\angle ABC =\angle BAC + \angle ACB = \angle BAC + \angle CAE $, $\angle BAE$ is a right angle and $A,E,D$ are colinear 
Clearly $|CE| = |AB|$ and $\angle AEC$ is a right angle.
Then $\angle CED $ is a right angle and also $|CE| = |CD|$, so $\angle ECD = 0 $ and $D$ and $E$ are the same point.
