Proposition: 1.43: Complementary Segments of Parallelograms

Euclid's Formulation

For any parallelogram, the complements of the parallelograms about the diagonal are equal to one another.

Let $ABCD$ be a parallelogram, and $AC$ its diagonal.
And let $EH$ and $FG$ be the parallelograms about $AC$, and $BK$ and $KD$ the so-called complements (about $AC$).
I say that the complement $BK$ is equal to the complement $KD$.

fig43e

Modern Formulation

Let $\boxdot{ABCD}$ be a parallelogram with a diagonal $\overline{AC}$ and let $K$ be any point on that diagonal. Then segments, which are parallel to the sides of the parallelogram and pass through $K$, (specifically $\overline{EF}$, $\overline{GH}$) divide $\boxdot{ABCD}$ into four smaller parallelograms: the two segments through which the diagonal does not pass ($\boxdot{EBGK}$, $\boxdot{HKFD}$) are called the complements of the other two ($\boxdot{AEKH}$, $\boxdot{KGCF}$). Moreover, the complements are equal in area: $\boxdot{EBGK}=\boxdot{HKFD}$.

Proofs: 1

Proofs: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18