# 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 $KD$. ### 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

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### References

#### Adapted from CC BY-SA 3.0 Sources:

1. Callahan, Daniel: "Euclid’s 'Elements' Redux" 2014