The Garden of Archimedes  A Museum for Mathematics

Pythagoras and his theorem

Parallelograms and trapeziums

	The theorem of Pythagoras can also be enunciated in a different form: the sum of the squares of the base and the altitude of a rectangle is equal to the square of the diagonal. In fact, the diagonal of a rectangle is the hypotenuse of the right-angled triangle having as cathets the base and the altitude. If we then take each square twice, we'll have that: the sum of the squares on the sides of a rectangle is equal to the sum of the squares on the diagonals.
	The same result is also valid for the non right-angled parallelogram. Take the parallelogram ABCD. When the theorem of Pythagoras is applied to the right-angled triangle BED, the square of diagonal BD is equal to the sum of the squares of ED and BE, which are coloured in green and yellow. Similarly, the square of the diagonal AC is equal to the red square plus the multicolored one. The sum of the areas of the squares of the diagonals is then equal to the one of the areas of the four squares drawn in the first figure. The second figure is obtained from the first by moving some parts without altering the overall area, and therefore the sum of the areas of the four squares of the first figure (which was equal to the sum of the squares of the diagonals) is equal to the one of the six squares of the second figure. On the other hand, again with the theorem of Pythagoras, the two green squares are equal to the square of the AB, and the two red ones to the square of side AC, and thus the sum of the areas of the six squares is equal to that one of the squares of the sides. We can conclude by saying that: in a parallelogram the sum of the areas of the squares of the diagonals is equal to the sum of the areas of the squares of the four sides.
	A similar result is also valid for trapeziums: the sum of the areas of the squares of the sides is equal to the sum of the areas of the squares of the diagonals, plus the square of the difference between the longer and the shorter base. In this case the best demonstration is the algebraic one. Referring to the following figure we have to demonstrate that: L2 + l2 + B2 + b2 = D2 + d2 + (B - b)2. We start by applying the theorem of Pythagoras to the triangles with sides h, q, L and h, p, l. We have: h2 + q2 =L2 h2 + p2 = l2 Similarly, by applying the theorem of Pythagoras to the triangles with sides h, D, b + q and h, d, b + p, the result is: D2 = (b + q)2 + h2 = b2 +q2 + 2bq + h2 = b2 +L2 + 2bq d2 = (b + p)2 + h2 = b2 +p2 + 2bp + h2 = b2 +l2 + 2bp and summing up D2 + d2 = L2 + l2 + 2b (b + p+ q) = L2 + l2 + 2bB. Incidentally, it can be observed that this last formula, translated into geometrical terms, says that: In a trapezium, the sum of the squares of the diagonals is equal to the sum of the squares of the non parallel sides, plus twice the rectangle of the bases. To obtain the wanted result, we use the formula (B - b)2 = B2 + b2 – 2Bb, or 2Bb = B2 + b2 – (B - b)2 , that inserted in the previous one, gives D2 + d2 = L2 + l2 + B2 + b2 – (B - b)2 , which is what we wanted to demonstrate.
	In the case of a right-angled trapezium, a very simple visual demonstration can be given. In the following figure, applying the theorem of Pythagoras to the triangles ABC and ABD, we see that the sum of the squares of the diagonals is equal to the four squares drawn. Among these, the yellow ones are the squares of the respective sides, while the square of side CD is obtained by adding the square DE to the green one, which is the difference of the parallel sides. If, then, to the squares of the diagonals we add that of the difference of the parallel sides, we obtain the sum of the squares of the sides.