Given two adjacent sides of a rectangle are equivalent, prove that the quadrilateral is a square.

Question

In Geometry class today, we were talking about quadrilaterals and the types of them. I was wondering that if, given a rectangle with two adjacent equal congruent sides, if that was enough to prove it was a square. I thought surely, it MUST be a square, because of... And there I was lost for a while. I flipped through my notes looking for some theorem that proved this in a step or two, and, finding none, I tried my own way of proving it. Did I prove it correctly? If not, how do I correct it? Finally, is there a shorter, more elegant way to execute it? (The below is the proof I have.)

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A               B
 ┌─┬────┼────┬─┐
 ├─┘         └─┤
 │             │
 │             ┼
 │             │
 ├─┐         ┌─┤
 └─┴─────────┴─┘
D               C

Conjecture: If $\overline{AB}\cong\overline{BC}$ and $ABCD$ is a rectangle, then $ABCD$ is a square.

Proof.

$$\begin{array}{cc|cc} \text{Statements}&&\text{Reasons}\\\hline \overline{DB}\cong\overline{DB}&&\text{Reflexive property of Congruency}\\\hline \angle ABD\cong\angle BDC&&\text{AIA* Theorem}\\ \angle DBC\cong\angle ADB\\\hline \triangle ABD\cong \triangle CBD&&\text{SAS** Theorem}\\\hline \overline{AD}\cong\overline{CB}&&\text{CPCTC***}\\ \overline{AB}\cong\overline{CD}\\\hline ABCD\text{ is a square}&&\text{Definition of Square}&\blacksquare \end{array}$$

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_{*Alternate Interior Angle}

_{**Side angle side}

_{***Congruent Parts of a Triangle are Congruent}

Answer 1

Given: ABCD is a rectangle, AB = BC

Prove: ABCD is a square

Step 1: AB = CD, BC = AD (property of rectangle: opposite sides are congruent)

Step 2: AB = AD, BC = CD (transitive property from Step 1 and Given)

Step 3: ABCD is a square (definition of square: rectangle where all sides are congruent)

QED