Teaching Pythagoras’ Theorem Visually
Could not be simpler!
There are many ways to demonstrate Pythagoras’ theorem. The following approach is simple, visual, uses basic geometry, and uses colors to make the explanation more evident.
The idea: Red + Blue = Yellow
Step 1. We slide the red square to the right:
Step 2. We slide the blue square up, and to the right, as shown:
The End Result: In the previous diagram, we can see three triangles, two blue and one red, that can replace the congruent (same size and shape) three yellow triangles.
Thus, the blue area plus the red area covers exactly
the area of the yellow square:
The Geometrical Demonstration
Our gray right angle triangle is marked as #1.
We oriented our triangle so that the longer side [b] is at the bottom and the hypotenuse [c] is at the top-right.
We want to show that the areas of the red square (a x a) and the blue square (b x b) add up to the area of the yellow square.
We call α the triangle’s smaller internal angle in point R.
Step 1. We first move our red square to the right, without changing its orientation, until its right bottom corner reaches point R. This creates a small right angle triangle near the top-left of the red square.
The length of this right angle triangle’s top side is b — a.
Its internal angle in point S is equal to α (the α angles are intern alternates of the square’s parallel horizontal edges intersected by the hypotenuse).
Step 2. This intermediate step is just to show that the distance between the top edge of the red square and point T is equal to the edge of the blue square (b) (See diagram below).
We temporarily move the blue square, without changing its orientation, to the position shown on the diagram below, with its bottom right corner in point S.
The red and blue squares are parallel to each other. Two construction lines can be drawn extending the horizontal edges of the blue square to the right.
The bottom construction line obviously goes through the top edge of the red square.
The top construction line will encounter point T and form a right angle triangle marked as #2. This is because we did not change the squares’ orientation and we have formed a second angle with the value α in point S. Consequently, triangles #1 and #2 are congruent, having two equal angles (α and the right angle) and an equal segment in between (b).
Step 3. We now move the blue square to the right until its left-top left corner is in point T (See diagram below) .
By the same reasoning as in Step 2, the right-most edge of the blue square now intersects point U: We created two other angles α, one in point T and another one in point U, and we created two new triangles marked as #2 congruent with each other and with our original #1.
The vertical edge of the triangle on the top-right has length a, because in the previous step we saw that triangle #2 was congruent with triangle #1 (with edges a, b, and c). Thus, the leftover segment UV has length b — a.
We can now see that the two triangles marked as #3 are congruent as they have two equal angles (α and the right angle), and an equal edge in between of length b — a.
The triangles marked with #4 are also congruent, as they have two equal angles and the same length edge in between, of length a.
The End. Our proof is completed.
To visualize it, we can change the color of the three yellow triangles with the color of their congruent blue and red triangles respectively (See the diagram below).
Thus, the area of the yellow square is exactly covered by blue and red, representing the areas of the two smaller squares.
Note. If our right angle triangle was isosceles (b = a), then the couple of
triangles #3 disappear and the last step of our procedure would lead to the simpler diagram below.
