Sufficient Conditions for a Rhombus

Sufficient Conditions for a Rhombus — Moosa Academy

A rhombus has perpendicular diagonals, angle-bisecting diagonals and four equal sides. Each of those properties also runs the other way: spot any one of them in a parallelogram and the shape must be a rhombus. These are the converses of the theorems you already know.

Theorem Perpendicular diagonals

If the diagonals of a parallelogram are perpendicular, the parallelogram is a rhombus.

If $JL \perp KM$ , then $JKLM$ is a rhombus.

Note the starting point: the shape must already be a parallelogram. Perpendicular diagonals alone, in an arbitrary quadrilateral, only give a kite.

Theorem A diagonal that bisects its corner angles

If a diagonal of a parallelogram bisects both of the angles it connects, the parallelogram is a rhombus.

In $WXYZ$ , if the diagonal splits each corner angle into two equal halves, then $WXYZ$ is a rhombus.

One diagonal is enough. Once it bisects its two corner angles, the other diagonal is forced to do the same.

Theorem Two consecutive sides equal

If two consecutive sides of a parallelogram are congruent, the parallelogram is a rhombus.

If $AB \cong BC$ , then $ABCD$ is a rhombus.

Two sides settle all four. Opposite sides of a parallelogram are already equal, so making one neighbouring pair equal forces every side to the same length.

Theorem Rectangle and rhombus together

A quadrilateral that is both a rectangle and a rhombus is a square.

A rhombus whose four angles are right angles is a square.

A rectangle whose four sides are congruent is a square.

The two statements say the same thing from opposite ends. A rectangle supplies the right angles, a rhombus supplies the equal sides, and a square is the shape that has both.

Example Testing a parallelogram

In parallelogram $JKLM$ the diagonals cross at $N$ , with $\angle JNK = 2x + 10$ and $\angle KNL = 4x - 6$ . Is it a rhombus?

For a rhombus the diagonals must be perpendicular, so every angle at

N

90^\circ

Test the first:

2x + 10 = 90

, giving

x = 40

Substitute into the second:

4(40) - 6 = 154^\circ

But

154^\circ \neq 90^\circ

, so the diagonals are not perpendicular

$JKLM$ is not a rhombus

The condition either holds throughout or not at all. Finding a value of $x$ that satisfies one angle proves nothing on its own — it has to satisfy the other as well.

Summary

These conditions are the converses of the rhombus properties.
Every one of them assumes the shape is already a parallelogram.
Perpendicular diagonals $\Rightarrow$ rhombus.
A diagonal bisecting both of its corner angles $\Rightarrow$ rhombus.
Two consecutive sides congruent $\Rightarrow$ rhombus.
Rectangle $+$ rhombus $=$ square: right angles from one, equal sides from the other.