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

 * PappusPythagorasExtension.png

From Parallelograms with Same Base and Same Height have Equal Area:
 * $ACDE = ACHR = ATUJ$

and:
 * $BCFG = BCHS = BIUT$

Hence the result.