# 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.

• What about setting $\theta$ as the angle $\angle C$, then $CD=AB\sin(\theta)=AB$ so $\theta=90°$ Commented May 26, 2015 at 18:17
• @Farnight No trig please (or as little as possible because I don't really know it that well). Can you translate that into a triangle proof/a proof without trig? Thanks Commented May 26, 2015 at 18:19
• Sorry. Geometry isn't my strong point. Commented May 26, 2015 at 18:21

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

• Wait, but then $ADD'$ was never actually a triangle right? That sounds a little fishy... you used properties of a triangle, when it turns out that $ADD'$ may not actually be a triangle. Commented May 26, 2015 at 18:15
• @soktinpk Alright, then I show that the assumption $D\ne D'$ is absurd. That's not fishy Commented May 26, 2015 at 18:17
• We know $D$ and $D'$ have to be collinear, but how do we know they are the same distance from $A$? It's "obvious", but isn't the point here to prove something obvious? Commented May 26, 2015 at 18:19
• @DavidK Sorry, mixed up $A$ and $C$, corrected now. We have $CD'=AB$ from the rectangle $ABCD'$ and $AB=CD$ is given Commented May 26, 2015 at 18:20
• Ah, makes much more sense now. (I think the first "angle at $D$" was meant to be "angle at $D\,'$".) Commented May 26, 2015 at 18:28

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.

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.

• This was my attempt at a quick answer; but I think Hagen von Eitzen's (corrected) answer, posted while I was writing this, is a little more elegant. Commented May 26, 2015 at 18:31