Equal Areas In Parallelogram à la Pythagoras
The applet below presents a generalization of the Pythagorean theorem due to W. J. Hazard (Am Math Monthly, v 36, n 1, 1929, 32-34).
Let parallelogram ABCD be inscribed into parallelogram MNPQ. Draw BK||MQ and AS||MN. Let the two intersect in Y. Then
Area(ABCD) = Area(QAYK) + Area(BNSY).
|What if applet does not run?|
The proof is a slight simplification of the published one. It proceeds in 4 steps. First, extend the lines as shown in the applet.
Then, the first step is to note that parallelograms ABCD and ABFX have equal bases and altitudes, hence equal areas (Euclid I.35 In fact, they are nicely equidecomposable.) For the same reason, parallelograms ABFX and YBFW also have equal areas. This is step 2. On step 3 observe that parallelograms SNFW and DTSP have equal areas. (This is because parallelograms DUCP and TENS are equal and points E, S, H are collinear. Euclid I.43 then implies equal areas of parallelograms SNFW and DTSP) Finally, parallelograms DTSP and QAYK are outright equal.