Lesson More proofs of the Pythagorean Theorem

More proofs of the Pythagorean Theorem

The proof of the Pythagorean theorem is given in the lesson The Pythagorean Theorem.
But this theorem is so remarkable that mathematicians developed tens of different proofs of the Theorem.
You may ask: why? and for what? The answer is - because it is interesting, and new proofs are beautiful and challenging.
In this lesson we provide couple of other proofs of the same theorem.

The Pythagorean Theorem

The sum of the areas of the squares constructed on the legs of a right triangle  
is equal to the area of the square constructed on its hypotenuse.

An equivalent formulation is:
If a and b are the lengths of two legs of the right triangle, and c is the length of its hypotenuse, then
.
We will prove this last statement.

Proof by rearrangement

Let us consider the square ABCD with the side length a+b as shown in Figure 1. Let us introduce points E, F, G and H in its sides by dividing each side in two intervals of the length a and b (Figure 1). Connecting point E and F, F and G, G and H, H and E by straight intervals, we get four right triangles, EBF, FCG, GDH and HAE.   These four triangles are congruent, because they are right triangles and have equal legs. Hense, the triangles have the same area. Let us denote by the common area value of all of these triangles. Since all these triangles are congruent, the quadrilateral EFGH has four equal sides. The angle HEF of the quadrilateral is the right angle, because it complements two other angles, AEH and FEB, to 180°, while the sum of these two angles, AEH and FEB, is equal to 90°. The other angles of the quadrilateral EFGH are right also by the similar reason. So, the quadrilateral EFGH is the square. Let us denote by the common value of the length of all sides of this square. For the entire area of the square ABCD, divided in four equal triangles plus the square, we can write .

Figure 1.

Now consider another subdivision (layout) of the same square ABCD (Figure 2). This subdivision is produced by the straight line   EG parallel to AD and BC, and by the straight line HF parallel to AB and CD, with the intersection point K. Segment AE, HK, DG, AH, EK, BF lengths are all equal to a, while segment EB, KF, GC, HD, KG, FC lengths are all equal to b. Diagonals HG and EF subdivide rectangles HKGD and EBFK in pairs of right triangles. Note that all these four triangles are congruent as right triangles having congruent legs. In addition, they all are congruent to triangles in Figure 1 by the same reason. Therefore, these four triangles have the same area . Having this, we can write for the entire area of the square ABCD in Figure 2 . By comparing this with the expression for the area of the square ABCD obtained above, we get , exactly what the theorem states.

Figure 2.

Proof using similar triangles

Let ABC be a right triangle with the legs of length a and b and the hypotenuse of length c (Figure 3). Let us draw an altitude CD from the right angle C to the hypotenuse AB, and let d and e be the length of segments the altitude divides the hypotenuse (AD=d, DB=e), while h be the length of the altitude (h=CD). Triangles ADC and ABC are similar, because they both are right triangles having common acute angle CAD. Therefore the proportion is valid, hence . Next, triangles DBC and ABC are similar, because they both are right triangles having common acute angle CBD. Therefore the proportion is valid, hence . Further, we have . Substituting expressions for d an e obtained above to this formula, we get . Multiplying both sides by gives . Note that . This follows from the proportion , which is valid due to similarity of triangles ADC and ABC. Thus, the equality is proved.

Figure 3. A right triangle,         its legs and hypotenuse