# Why the Area of a Rhombus is Half the Product of its Diagonals

A rhombus is a parallelogram with four congruent sides. Since it is a parallelogram, it has also all the properties of a parallelogram. One of these properties is that the diagonals bisect each other. That is, they divide each other into two equal parts.

Another property of a rhombus is that the diagonals are perpendicular. So, summarizing all the properties above, if we have rhombus $ABCD$, then,

$\overline{AB} \cong \overline{BC} \cong \overline{CD} \cong \overline{DA}$.

and

$\overline{AC} \perp \overline{BD}$.

From the figure below, it is clear that if we divide the parallelogram into two triangles, $ABC$ and $ACD$,  $\overline{BE}$ and $\overline{DE}$ are their altitudes because they are perpendicular to $\overline{AC}$.

To find the area of parallelogram $ABCD$, first we find the area of triangle $ABC$, and then multiply it by 2. This is because triangle $ABC$ is congruent to triangle $CDA$ by SSS congruence (can you see why?).

Now, we know that the area of a triangle is 1/2 the product of the base and the height, so

area of triangle $ABC = \frac{1}{2} (AC)(BE)$.

But we also know that $BE$ is half of $BD$, so

Area of $ABC = \frac{1}{2}(AC) (\frac{1}{2} BD) = \frac{1}{4} (AC)(BD)$.

Now, to get the area of ABCD, we multiply the area of ABC by 2. That is,

Area of parallelogram $ABCD = 2 \left (\displaystyle \frac{1}{4}(AC)(BD) \right ) = \frac{1}{2} (AC)(BD)$

This is what we want to prove: the area of parallelogram $ABCD$ is half the product of diagonals $AC$ and $BD$.