Theorem: Egyptian Formula for Area of Quadrilateral

Let $\Box ABCD$ be a quadrilateral.
Let the sides of $\Box ABCD$ be $a$, $b$, $c$ and $d$ such that $a$ is opposite $c$ and $b$ is opposite $d$.
Then the area of $\Box ABCD$ can be approximated by:
:$\map \Area {\Box ABCD} \approx \dfrac {a + c} 2 \times \dfrac {b + d} 2$
The closer $\Box ABCD$ is to a rectangle, the better the approximation.

Proof:
thumbA [[Definition:Quadrilateralquadrilateral $\Box ABCD$ with side lengths $a = \overline {AB}$, $b = \overline {BC}$, $c = \overline {CD}$, and $d = \overline {DA}$. The point $C'$ is positioned so that $\Box ABC'D$ is approximately a rectangle if $\angle DAB$ is approximately a right angle.]]
We have a quadrilateral $\Box ABCD$ with side lengths
:$a = \overline {AB}$
:$b = \overline {BC}$
:$c = \overline {CD}$
:$d = \overline {DA}$
Let $\angle DAB$ be approximated as a right angle.
Extend a line from $B$ parallel to $\overline {AD}$ towards $\overline {CD}$.
Extend a line from $D$ parallel to $\overline {AB}$ towards $\overline {BC}$.
The point where these two lines intersect is $C'$.
The total area of $\Box ABCD$ is equal to the area of the quadrilateral $\Box ABC'D$ and the two triangles $\triangle BC'C$ and $\triangle DC'C$.
:$\map \Area {\Box ABCD} = \map \Area {\Box ABC'D} + \map \Area {\triangle BC'C} + \map \Area {\triangle DC'C}$
By construction, the quadrilateral $\Box ABC'D$ has opposite sides parallel, and so is a parallelogram.
Parallelogram with One Right Angle is Rectangle, so $\Box ABC'D$ is a rectangle.
By Area of Rectangle, $\map \Area {\Box ABC'D} = a d$.
$\triangle BC'C$ has a base of $d$.
$\triangle BC'C$ has a height approximated by $c - a$.
So, by Area of Triangle in Terms of Side and Altitude,
:$\map \Area {\triangle BC'C} \approx \frac 1 2 d \paren {c - a}$.
Note that, if $c < a$, then the resulting area for $\triangle BC'C$ would come out negative, correctly subtracting its area from $\Box ABC'D$.
Mutatis mutandis,
:$\map \Area {\triangle DC'C} \approx \frac 1 2 a \paren {b - d}$.
Therefore,
{{begin-eqn}}
{{eqn | l = \map \Area {\Box ABCD}
      | r = \map \Area {\Box ABC'D} + \map \Area {\triangle BC'C} + \map \Area {\triangle DC'C}
}}
{{eqn | o = \approx
      | r = a d + \frac 1 2 d \paren {c - a} + \frac 1 2 a \paren {b - d}
}}
{{eqn | r = \frac 1 2 \paren {a b + c d}
}}
{{end-eqn}}
The choice of starting this calculation with angle $\angle DAB$ was arbitrary and ignores the other three angles.
To account for all angles of the quadrilateral, repeat the above procedure with every angle and take the mean of the resulting areas.
To be specific, the procedure for $\angle ABC$ would be
:$(1):$ Let $\angle ABC$ be approximated as a right angle.
:$(2):$ Extend a line from $C$ parallel to $\overline {AB}$ towards $\overline {AD}$.
:$(3):$ Extend a line from $A$ parallel to $\overline {BC}$ towards $\overline {CD}$.
:$(4):$ The two lines intersect at $D'$.
:$(5):$ $\map \Area {\Box BCD'A} = a b$
:$(6):$ $\map \Area {\triangle CD'D} \approx \frac 1 2 a \paren {d - b}$
:$(7):$ $\map \Area {\triangle AD'D} \approx \frac 1 2 b \paren {c - a}$
:$(8):$ $\map \Area {\Box ABCD} \approx \frac 1 2 \paren{a d + b c}$
Thus, the area found using all four angles of $\Box ABCD$ is
{{begin-eqn}}
{{eqn | l = \map \Area {\Box ABCD}
      | o = \approx
      | r = \frac 1 4 \paren {\frac 1 2 \paren {a b + c d} + \frac 1 2 \paren {a d + b c} + \frac 1 2 \paren {a b + c d} + \frac 1 2 \paren {a d + b c} }
}}
{{eqn | r = \frac 1 4 \paren {a b + c d + a d + b c}
}}
{{eqn | r = \frac {a + c} 2 \times \frac {b + d} 2
}}
{{end-eqn}}
{{qed}}