Pythagoras's Theorem for Parallelograms
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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.)