A dissection proof of the law of cosines, replacing Cuoco-McConnell's rectangles with congruent triangles
aa r X i v : . [ m a t h . HO ] J un A dissection proof of the law of cosines,replacing Cuoco-McConnell’s rectangles with congruent triangles.Martin Celli
March 31st 2018Departamento de Matem´aticasUniversidad Aut´onoma Metropolitana-IztapalapaAv. San Rafael Atlixco, 186. Col. Vicentina. Del. Iztapalapa. CP 09340. Mexico City.E-mail: [email protected] up the challenge McConnell laid down at the end of his proof of the law ofcosines ([6]), we give a completely visual dissection proof of this theorem, which applies toany triangle. In order to avoid the trigonometric expressions of Cuoco-McConnell’s proof,we replaced the equal-area rectangles with congruent triangles. As a matter of fact, tri-gonometric expressions are implicitly based on the similarity of two right triangles with acommon non-right angle. So they are conceptually less simple than our congruent trian-gles which are, moreover, easy to visualize. This makes our proof the only dissection proofand the simplest proof of its family, and thus one of the best options for a course of geometry.Cuoco-McConnell’s proof was published in [6], [3], [5], [8]. In [8], the proof is writtenin an algebraic way, whereas in [6], [3], [5], it is written in an equivalent geometric wayinvolving three pairs of rectangles with the same area. A similar proof, involving two pairsof rectangles with the same area, can be found in [7]. Unfortunately, in these references, theareas of all the rectangles had to be expressed as trigonometric functions, in order to showthat they were pairwise equal. In the particular case of a right triangle (where a pair ofrectangles have zero area), and as in our proof, Euclid had already managed to avoid thisby considering, for every pair of rectangles, a pair of congruent triangles with half the areaof one of the rectangles ([4]). A generalization of Euclid’s proof to the law of cosines canbe found in [2] (where every triangle is replaced with a parallelogram made of this triangleand its symmetric). Each triangle of Euclid, as well as each triangle of our proof, has twosides in common with the triangle
ABC of the theorem. However, denoting by x the angleformed by these sides, the corresponding angle in Euclid’s triangle is π/ x , whereas thecorresponding oriented angle in our triangle is π/ − x . As a consequence of this, and unlikeEuclid’s triangles and the parallelograms of [2], two triangles of a same pair of our proof donot have intersection. This helps us in finding a visual dissection argument (cutting eachrectangle into a set of congruent pieces) in order to prove that the area of the triangle is halfthe area of the corresponding rectangle. In both our proof and [2], trigonometry is only usedto compute the inevitable area of the triangles/parallelograms corresponding to the term2 CA.CB cos(
BCA ) of the identity.When one of the points A ′ , B ′ , C ′ of our proof (symmetric of A , B , C with respect to BC , CA , AB ) is located outside its square, some areas need to be interpreted in an oriented1ense, and counted as negative. This already happened in [6], [3], [5], when the triangle wasobtuse. Here, it can also happen when an altitude of the triangle ABC is longer than itsperpendicular side. We give the proof in the acute case, when the altitude passing through B is longer than AC (thus, the point B ′ is outside its square), the other cases can be studiedin a similar way.It is worth mentioning that, in the particular case where the three points A ′ , B ′ , C ′ are located inside their squares, our proof becomes equivalent to Anderson’s ([1]), where bothsmall triangles adjacent to a same blue/green/red triangle of the figure are cut and pasted,so that the three triangles (or equivalently, the blue/green/red triangle and its symmetric)make a parallelogram. Thus, applying the same transformation (replacing a triangle with aparallelogram made of it and its symmetric), we can obtain the parallelograms of [2] fromEuclid’s triangles, and Anderson’s parallelograms from the triangles of the proof of this note.As every triangle of this proof has two sides in common with a triangle of Euclid, with acorresponding angle π − α = π/ − x instead of α = π/ x , every parallelogram of Andersonand its corresponding parallelogram of [2] are congruent.Here is our proof: BC + CA − AB = [red square] + [green square] − [blue square]= (2 A + 2 A + 2 A + 2 A ) + ((2 B − B ) + (2 B − B )) − (2 C + 2 C + 2 C + 2 C )= 2(( A + A ) + ( A + A ) + ( B − B ) + ( B − B ) − ( C + C ) − ( C + C ))2 2([green triangle]+[blue triangle]+[blue triangle]+[red triangle] − [red triangle] − [green triangle])= 4[blue triangle] = 2 CA.CB sin( π/ − BCA ) = 2
CA.CB cos(