“_In any right-angled triangle, the square which is described
upon the side opposite to the right-angle is equal to the squares
described upon the sides which contain the right-angle._”
Here is a simple way of proving this proposition. Although perhaps
not exactly scholastic, it is none the less interesting.
Draw an exact square, whose sides measure 7 in.; then divide it into
49 square inches. Having done this, cut the figure in following the
big lines as shown by Fig 1. It will be observed that C is a complete
square, and that A and B will form a square: but as D is 1 in. short
of being a square, it is necessary to cut a square inch and add it on.
[Illustration: Fig. 1.]
[Illustration: Fig. 2.]
Then construct a right-angled triangle as shown by Figure 2.
We then see that the sum of the two small squares is equivalent to
the large square.
D contains 9 small squares.
A & B do. 16 do.
--
25
And as we see that C has 25 small squares, it is thus proved that the
sum of the squares upon the sides which contain the right angle are
equal to the squares upon the side opposite the right angle.
_Q.E.D._