Elegant proof of Pythagoras` theorem
TO ASSESS the teaching power of Krishna Tirtha's book, take an example from geometry. Schoolchildren are familiar with Pythagoras' theorem that states the square on the hypotenuse of a right-angled triangle is equal to the sum of the squares on the other two sides.
This theorem is proved in standard geometry textbooks with drawings of perpendiculars and triangles and in a manner that is difficult for the young student to understand and remember. So they generally rely on memorising the proof.
But in Krishna Tirtha's method, all that you have to do is to draw a square within a square to prove Pythagoras' theorem.
In a square ABCD, each side is equal to a. Mark the points E, F, G and H on each side of the square so that AE, BF, CG and DH are all equal to m and BE, CF, DG and AH are equal to n. Connect the points to form the square EFGH in which each side is equal to h.
Clearly, a = m + n.
The area of ABCD is a2, or (m+n)2 = m2 + n2 + 2mn. But the area of the larger square is also the area of the smaller square plus the areas of the four identical right-angled triangles EBF, FCG, GDH and HAE. The area of the smaller square is h2.
The area of a triangle is half the base multiplied by altitude. In the triangle EBF, for instance, the base is EB, which is equal to "n" and the altitude is BF, which is equal to "m". FE, which forms one side of the square EFGH and is equal to "h", is also the hypotenuse of the triangle EBF. The area of each triangle is (1/2)mn. The area of four such triangles is 2mn.
By this argument, h2+ 2mn (area of small square + area of 4 triangles) = m2 +n2 + 2mn (area of big square).
This leaves h2 (square on the hypotenuse) = m2 + n2 (sum of squares on the other sides of the triangle).
This is precisely the substance of the Pythagoras theorem. Can anything be simpler or more elegant?