Is it possible to prove a quadrilateral a parallelogram with two consecutive and two opposite congruent sides?

This is a problem that came up during my geometry class, and, though logical, neither me or my teacher could come up with why.

Consider quadrilateral ABCD, listing clockwise, with A in the top left, depending on how you look at it.

Angle DAB is a right angle. Angle CDA is a right angle. Segment AD is congruent to segment BC.

Due to the fact that two lines perpendicular to a third are parallel, Segment AB is || to Segment CD.

From my thinking, since the opposite sides are parallel, and the other pair is congruent, and there are two consecutive right angles, the only possible way Segment BC could connect points B and C while being congruent to segment AD is if it meets segments AB and CD at right angles.

Is there a theorem that proves this figure is a parallelogram, or is my thinking simply a postulate? Is it possible to disprove my thinking?

• That is what I cannot prove. DAB is a right angle, and so is CDA, so they are both perpendicular to segment AD, and therefore parallel, because of the consecutive interior angle converse theorem. – mrdorkface Feb 18 '16 at 15:50
• sorry i didn't read your post properly. Thanks. – Max Payne Feb 18 '16 at 16:12

There are lots of ways to show this. The most straightforward one is by drawing perpendicular from $B$ to line $CD$. Let $E$ be the intersection. If $E\neq C$ then $\triangle BEC$ is a right triangle, with $\angle E=90^\circ$ and $|AD|=|BE|=|BC|$ which is impossible.
• @mrdorkface, $E$ is the intersection of line $CD$ (not segment $[CD]$) and line, going through $B$, perpendicular with line $CD$. What i mean is draw perpendicular line from $B$ to line $CD$; and $E$ is the intersection of these two lines. – Alistair Feb 18 '16 at 16:10