Parallelograms with Equal Base and Same Height have Equal Area

Theorem

Parallelograms which have equal bases and in the same parallels are equal to one another.

Proof

Let $ABCD$ and $EFGH$ be parallelograms with equal bases $$ and $FG$, and in the same parallels $AF$ and $BC$.

Join $BE$ and $CH$.

We have $BC = FG$ and $FG = EH$ so by Common Notion 1 we have that $BC = EH$.

But $EB$ and $HC$ join them.

So by Lines Joining Equal and Parallel Straight Lines, $EB$ and $HC$ are equal and parallel.

So $EBCH$ is a parallelogram.

So, by Parallelograms with Same Base and Same Height have Equal Area, the area of $EBCH$ equals the area of $ABCD$.

For the same reason, the area of $EBCH$ equals the area of $EFGH$.

So by Common Notion 1, the area of $ABCD$ equals the area of $EFGH$.

$\blacksquare$

Historical Note

This is Proposition 36 of Book I of Euclid's The Elements.