The rhombus

Definition: a rhombus is a parallelogram with two equal adjacent sides.

A rhombus is a parallelogram, so it has all the properties of a parallelogram: opposite sides parallel and equal, opposite angles equal, and diagonals that bisect each other.

Theorem: all the sides of a rhombus are equal.

In the rhombus $\displaystyle ABCD$, $\displaystyle {AB}={AD}$.

• $\displaystyle {AB}={DC}$ and $\displaystyle {AD}={BC}$, since the opposite sides of a parallelogram are equal;
• so all four sides are equal: $\displaystyle {AB}={BC}={CD}={DA}$.

Theorem: the diagonals of a rhombus are perpendicular, and each diagonal bisects the angles at its ends.

In the rhombus $\displaystyle ABCD$ the diagonals meet at the point $\displaystyle O$:

• $\displaystyle AC\perp BD$;
• $\displaystyle AC$ bisects the angles $\displaystyle A$ and $\displaystyle C$, and $\displaystyle BD$ bisects the angles $\displaystyle B$ and $\displaystyle D$.

Given: $\displaystyle ABCD$ is a rhombus, and its diagonals meet at the point $\displaystyle O$.
Prove: $\displaystyle AC\perp BD$, and $\displaystyle AC$ bisects the angle $\displaystyle A$.

The proof below draws both from one congruence – of the triangles $\displaystyle ABO$ and $\displaystyle ADO$.

In the rhombus $\displaystyle ABCD$, $\displaystyle {AC}={16}$ and $\displaystyle {BD}={12}$.
The diagonals meet at the point $\displaystyle O$.

• The diagonals bisect each other, so $\displaystyle {AO}={8}$ and $\displaystyle {BO}={6}$;
• the diagonals are perpendicular, so in triangle $\displaystyle AOB$ the angle $\displaystyle O$ is right;
• by Pythagoras’ theorem $\displaystyle {AB^2}={8^2+6^2}={100}$, so $\displaystyle {AB}={10}$;
• all the sides are equal, so the perimeter of the rhombus is $\displaystyle {4\cdot10}={40}$.

In the rhombus $\displaystyle ABCD$, $\displaystyle {\angle ABC}={70^{\circ}}$.
The diagonals meet at the point $\displaystyle O$.

• $\displaystyle BD$ bisects the angle $\displaystyle B$, so $\displaystyle {\angle ABD}={35^{\circ}}$;
• in triangle $\displaystyle AOB$ the angle $\displaystyle O$ is right, so $\displaystyle {\angle BAO}={180^{\circ}-90^{\circ}-35^{\circ}}={55^{\circ}}$.

Three converse theorems prove that a quadrilateral is a rhombus:

Theorem: a parallelogram whose diagonals are perpendicular is a rhombus.

Theorem: a parallelogram with a diagonal that bisects one of its angles is a rhombus.

Theorem: a quadrilateral whose four sides are all equal is a rhombus.

Given: $\displaystyle ABCD$ is a parallelogram, and its diagonals are perpendicular and meet at the point $\displaystyle O$.
Prove: $\displaystyle ABCD$ is a rhombus.

The proof below shows that two adjacent sides, $\displaystyle AB$ and $\displaystyle CB$, are equal: they are sides of the triangles $\displaystyle AOB$ and $\displaystyle COB$, which are congruent.

In the parallelogram $\displaystyle ABCD$, $\displaystyle {\angle BAC}={\angle DAC}$: the diagonal $\displaystyle AC$ bisects the angle $\displaystyle A$.

• $\displaystyle {\angle DAC}={\angle BCA}$, since they are alternate angles between the parallel lines $\displaystyle AD$ and $\displaystyle BC$;
• so $\displaystyle {\angle BAC}={\angle BCA}$, and the triangle $\displaystyle ABC$ is isosceles: $\displaystyle {AB}={CB}$;
• a parallelogram with two equal adjacent sides is a rhombus.

In the quadrilateral $\displaystyle ABCD$, $\displaystyle {AB}={BC}={CD}={DA}$.

• Both pairs of opposite sides are equal, so $\displaystyle ABCD$ is a parallelogram;
• it has two equal adjacent sides, so it is a rhombus.

A quadrilateral is a rhombus if one of these holds:

• it is a parallelogram with two equal adjacent sides (the definition);
• all four of its sides are equal;
• it is a parallelogram whose diagonals are perpendicular;
• it is a parallelogram with a diagonal that bisects one of its angles.

$\displaystyle ABCD$ is a rhombus, and its diagonals meet at the point $\displaystyle O$.
Work it out and give the reasons.

$\displaystyle {AC}={10}$ and $\displaystyle {BD}={24}$.
Find the length of the rhombus’s side and its perimeter.

$\displaystyle {\angle ABC}={110^{\circ}}$.
Find the angles $\displaystyle BAC$ and $\displaystyle ABD$.

$\displaystyle {\angle BAD}={60^{\circ}}$ and $\displaystyle {AB}={6}$.
Find $\displaystyle BD$.

$\displaystyle ABCD$ is a rhombus, and the point $\displaystyle E$ is on the diagonal $\displaystyle AC$.
Prove: $\displaystyle {EB}={ED}$.

$\displaystyle ABCD$ is a kite: $\displaystyle {AB}={AD}$ and $\displaystyle {CB}={CD}$.
Also, $\displaystyle AB\parallel DC$.
Prove: $\displaystyle ABCD$ is a rhombus.