Showing a parallelogram is a rhombus using perpendicular diagonals

Key concept: A vector proof links the diagonal condition to equal side lengths through dot products.

Step by step

Set up the vectors

Let the adjacent sides of parallelogram $OPQR$ be

$\overrightarrow{OP}=\mathbf{p}, \qquad \overrightarrow{OR}=\mathbf{r}.$

Then the fourth vertex is $Q$, and the diagonals are

$\overrightarrow{OQ}=\mathbf{p}+\mathbf{r}, \qquad \overrightarrow{PR}=\mathbf{r}-\mathbf{p}.$

The problem states that these diagonals are perpendicular, so

$(\mathbf{p}+\mathbf{r})\cdot(\mathbf{r}-\mathbf{p})=0.$

Use the perpendicularity condition

Expand the dot product:

$\mathbf{p}\cdot\mathbf{r}-\mathbf{p}\cdot\mathbf{p}+\mathbf{r}\cdot\mathbf{r}-\mathbf{r}\cdot\mathbf{p}=0.$

Because dot product is commutative, $\mathbf{p}\cdot\mathbf{r}=\mathbf{r}\cdot\mathbf{p}$. These two terms cancel, leaving

$-|\mathbf{p}|^2+|\mathbf{r}|^2=0.$

Hence

$|\mathbf{p}|^2=|\mathbf{r}|^2,$

so $|\mathbf{p}|=|\mathbf{r}|$.

Conclude that the parallelogram is a rhombus

In a parallelogram, opposite sides are equal and parallel, so the side lengths are

$OP=QR=|\mathbf{p}|, \qquad OR=PQ=|\mathbf{r}|.$

Since $|\mathbf{p}|=|\mathbf{r}|$, all four sides are equal. A parallelogram with all sides equal is a rhombus.

Why this method works

The diagonals of a parallelogram are expressed naturally using the side vectors. The perpendicular diagonal condition gives a dot product equation, and that equation forces the two adjacent side lengths to be equal. This is a neat algebraic route to a geometric result.

For an exam answer, the key sequence is: write the diagonals as $\mathbf{p}+\mathbf{r}$ and $\mathbf{r}-\mathbf{p}$, set their dot product to zero, simplify, and conclude that the side lengths match.

Common mistake to avoid

Do not assume that perpendicular diagonals alone automatically mean a rhombus without showing the vector relation. The proof must connect the perpendicularity of the diagonals to equality of adjacent sides. Also, keep track of the sign in $\overrightarrow{PR}=\mathbf{r}-\mathbf{p}$; reversing it is fine, but the expansion must be consistent.

Pitfall alert

Students often stop after stating that a parallelogram with perpendicular diagonals is a rhombus, but that is the conclusion, not the proof. The exam marker wants to see the vector setup and the dot product calculation. Another error is writing the diagonals with the wrong direction, which can flip a sign and make the algebra look messy. The sign does not matter if you are consistent, but the final simplification must still show that the two adjacent side lengths are equal.

Try different conditions

If the question changes to "show that if a parallelogram has equal adjacent side lengths, then its diagonals are perpendicular," the proof runs in reverse. Start with $|\mathbf{p}|=|\mathbf{r}|$, then compute $(\mathbf{p}+\mathbf{r})\cdot(\mathbf{r}-\mathbf{p})$. The middle terms cancel and the result becomes $|\mathbf{r}|^2-|\mathbf{p}|^2=0$, so the diagonals are perpendicular. This is the converse statement and uses the same vector identities with the logic reversed.

Further reading

diagonals of a parallelogram, dot product, rhombus condition