In Figure 3, below, right triangle *ABC* with right angle at *B* is given with side lengths \(a, b,\) and \(c\) as shown. Rectangles are constructed on each side so that the width of the rectangles is the side length multiplied by some positive constant \(k,\) \(k\not=1\) (in Figure 3, \(k\) is about \(0.4\)). Can rectangle *AJHC* be divided up, or partitioned, in a natural way to illustrate the equation \[ka^2 + kb^2 = kc^2\] geometrically? In particular, can we proceed much as Euclid did for squares?

**Figure 3: **Similar rectangles on the sides of a right triangle. The Pythagorean cut, *LK,* is shown in red.

Following Euclid's proof of the Pythagorean Theorem, let *BK* be perpendicular to *JH* at *K* and intersect *AC* at *L*; draw construction lines *DC* and *BJ*, shown in green in Figure 3, above. We notice two things right away, one the same and one different from the situation in I.47 in which squares are drawn on the three sides. First, since triangle *DAC* and rectangle *DABE* share the same base *DA* and are in the same parallels with altitude *AB*, the area of rectangle *DABE* is twice the area of triangle *DAC*. Similarly, the area of rectangle *AJKL* is twice the area of triangle *JAB*. And second, although triangles *DAC* and *JAB* have a pair of congruent angles (*JAB* and *DAC*), the triangles are not congruent, since *DA* = *JA* only if \(a = c,\) which is impossible since \(c\) is the length of the hypotenuse, and *AJ* = *AC* only if \(k =1,\) which was not an option.

Even though triangle *DAC* is not congruent to triangle *JAB*, how do their areas compare? Triangle *DAC*, with base *DA* \(= ka\) and altitude *AB* \(= a,\) has area \({\frac{1}{2}}ka^2.\) Since the right triangles *ALB* and *ABC* are similar, we have \({\frac{AL}{a}}={\frac{a}{c}}.\) Hence, \(c\,AL = a^2,\) so that the area of triangle *JAB* is

\[{\frac{1}{2}}kc\,AL ={\frac{1}{2}}ka^2.\]

Thus, even though the two triangles are not congruent,

Area of triangle *DAC* = Area of triangle *JAB,*

so that

Area of rectangle *DABE* = Area of rectangle *AJKL*.

Similarly, we have

Area of rectangle *BCGF* = Area of rectangle *KHCL*,

so that

Area of rectangle *ACHJ* = Area of rectangle *BCGF* + area of rectangle *DABE*.

As in Euclid I.47, if point *L* is the foot of the perpendicular from *B* to *AC*, segment *LK*, perpendicular to hypotenuse *AC* (and parallel to rectangle side *AJ*), gives the natural partition of rectangle *ACHJ*. Thus, *LK*, shown in red in Figure 3, above, is the Pythagorean cut of rectangle *ACHJ*. A GeoGebra applet showing the Pythagorean cut for similar rectangles is given in Figure 4, below.

**Figure 4: **Pythagorean cut for similar rectangles

Please note that if similar parallelograms, rather than similar rectangles, were constructed on the sides of right triangle *ABC*, with point *L* determined as before and *LK* again the segment parallel to side *AJ* (see Figure 5, below), then the Pythagorean relationship among the areas of these parallelograms is still valid. That is,

Area of parallelogram *AJKL* = Area of parallelogram *EDAB*

and

Area of parallelogram *LKHC* = Area of parallelogram *BCGF*,

so that

Area of parallelogram *EDAB* + area of parallelogram *BCGF* = Area of parallelogram *AJHC*.

Again, *LK* is the Pythagorean cut for parallelogram *ACHJ*.

**Figure 5: **Similar parallelograms on the sides of a right triangle. The Pythagorean cut, *LK,* is shown in red.