Diagonals of a Rhombus

The diagonals of a rhombus have $3$ properties that we must remember.

The diagonals of a rhombus intersect. (not only do they intersect, but they do so right at the midpoint of each one).

When $ABCD$ is a rhombus
then:
$AE=CE$
$DE=BE$

The diagonals of a rhombus are perpendicular, they form a right angle of $90^o$ degrees.

When $ABCD$ is a rhombus
then:
$∢AED=∢AEB=∢DEC=∢CEB=90^o$

The diagonals of a rhombus bisect the angles of the rhombus.

When $ABCD$ is a rhombus
then:

$∢A1=∢A2$
$∢B1=∢B2$
$∢C1=∢C2$
$∢D1=∢D2$

(where A1 and A2 are the two angles formed when diagonal AC bisects vertex angle A, and similarly for the other vertices)

Observe:
These three statements are properties of the diagonals of a rhombus that, in case you had a rhombus in front of you, you should not prove them, but simply use them.
Anyway, to help you understand the logic behind them, we will demonstrate the properties below.

Diagonals of a rhombus

Given: $ABCD$ rhombus

To prove:
The diagonals of a rhombus intersect, bisect each other, are perpendicular, and bisect the vertex angles.

Solution:

Proving diagonals intersect

Since $AB∥DC$  and $AD∥BC$ the rhombus is also a parallelogram.
One of the properties of the parallelogram is that its diagonals intersect.
Therefore, diagonals AC and BD intersect at point E.
Thus, we have already proven the first property.

Proving the four triangles are congruent

Now we can prove that all triangles created from thediagonals are congruent.
All sides are equal: AB = BC = CD = DA
Let's see it clearly in the illustration:

Each triangle shares the intersection point E, therefore each triangle composed of a blue side, greenside and a red side. The triangles are congruent by SSS, (Side-Side-Side) congruence theorem: $△AEB ≅ △CED and △BEC ≅ △DEA$.
Therefore, all corresponding angles are equal.

Proving diagonals bisect each other

From the congruent triangles, corresponding sides are equal:

• $AE = CE$ (diagonal $AC$ is bisected at $E$)
• $BE = DE$ (diagonal $BD$ is bisected at $E$)

Proving diagonals are perpendicular

From the congruent triangles, corresponding angles are equal:

• $∠AEB = ∠CED$ (vertical angles)
• $∠BEC = ∠DEA$ (vertical angles)
• Since $∠AEB + ∠BEC = 180°$ (straight line) and $∠AEB = ∠BEC$ (from congruence), each angle must be $90°$
• Therefore:$∠AEB = ∠BEC = ∠CED = ∠DEA = 90°$

Proving diagonals bisect vertex angles

From triangle congruence, we can show that each diagonal splits the vertex angles equally:

• $∠DAE = ∠BAE$ (diagonal $AC$ bisects $∠A$)
• $∠ABE = ∠CBE$ (diagonal $BD$ bisects $∠B$)
• And similarly for vertices $C$ and $D$

The $2$ corresponding angles are also adjacent.
For the angles to be corresponding and also equal they must be right angles. Consequently, the diagonals are also perpendicular – The second property.

Useful Information:
We can deduce the area of the rhombus based on its diagonals!

Multiply the diagonals, divide by $2$ and we get the area of the rhombus.
Let's see it in the formula:

$\frac{product~of~the~diagonals}{2}=area~of~rhombus$

Given a rhombus $ABCD$

$AC=4$
and the area of the rhombus is equal to $40$

Find:
the length of the diagonal  $DB$

Solution:

Let's place in the formula
when $DB=X$
$\frac{X\cdot4}{2}=40$

Solve:
$X\cdot4=80$
$X=20$

Therefore, the diagonal $DB$ measures $20$.

Watch out, don't miss it!

You might sometime be asked if the diagonals of a rhombus are of the same length.
The answer is no.
The lengths of the diagonals of a rhombus are not equal.

Great! Now you know everything about the diagonals of a rhombus and you will be able to use some of its properties when it suits you.