Pythagoras's Theorem for Parallelograms

Theorem

Let $\triangle ABC$ be a triangle.

Let $ACDE$ and $BCFG$ be parallelograms constructed on the sides $AC$ and $BC$ of $\triangle ABC$.

Let $DE$ and $FG$ be produced to intersect at $H$.

Let $AJ$ and $BI$ be constructed on $A$ and $B$ parallel to and equal to $HC$.

Then the area of the parallelogram $ABIJ$ equals the sum of the areas of the parallelograms $ACDE$ and $BCFG$.

Proof

From Parallelograms with Same Base and Same Height have Equal Area:

$ACDE = ACHR = ATUJ$

and:

$BCFG = BCHS = BIUT$

Hence the result.

$\blacksquare$

Historical Note

Pythagoras's Theorem was extended by Pappus of Alexandria in this interesting manner of which Pythagoras's Theorem is a special case.

Sources

• 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {A}.8$: Pappus (fourth century A.D.)