Now we will discuss an interesting proof of the Pythagoras Theorem using the concepts we have covered in this chapter. Consider \(\Delta ABC\) which is right-angled at A:

Triangle

We need to show that AB2 + AC2 = BC2. We begin by drawing squares on each of the three sides of the triangle:

Squares and Triangles

Thus, effectively what we need to show is that:

area(Square ABGF) + area(Square CAED) = area(Square BCIH)

Do the following construction: join G to C, A to H, and draw a line through A perpendicular to BC, as show

Squares and Triangles with intersecting lines 

We note that \(\Delta GBC\) ≡ \(\Delta ABH\), because:

  1.  GB = AB (sides of the same square)
  2.  BC = BH (sides of the same square)
  3.  \(\angle GBC\) = 900 + \(\angle ABC\) = \(\angle ABH\)

Thus, the SAS congruence criterion applies. Since the two triangles are congruent, they will have the same area. Now comes the important part. Note that following:

  1. \(\Delta GBC\) and square GBAF are on the same base GB and between the same parallels, which means that area(GBAF) = 2 × area(\(\Delta GBC\)).

  2. \(\Delta ABH\) and rectangle BHKJ are on the same base BH and between the same parallels, which gives area(BHJK) = 2 × area(\(\Delta ABH\)).

From these two observations, we conclude that area(ABGF) = area(BHKJ). This is highlighted in the figure below:

Square and Rectangle with same areas 

Similarly, we can conclude that area(CAED) = area(CIKJ), as highlighted below:

Square and Rectangle with same areas

Clearly, if we add the areas of two smaller squares, we get the area of the larger square:

area(Square ABGF) + area(Square CAED) = area(Square BCIH)