Proposition XLVII. Theorem.
|(202)||In a right angled triangle (A B C) the square on the hypotenuse (A C) is equal to the sum of the squares of the sides (A B and C B).|
On the sides A B, A C, and B C describe the squares A X, A F, and B I, and draw B E parallel to either C F or A D, and join B F and A I.
Because the angles I C B and A C F are equal, if B C A be added to both, the angles I C A and B C F are equal, and the sides I C, C A are equal to the sides B C, C F, therefore the triangles I C A and B C F are equal (IV); by A Z is parallel to C I, therefore the parallelogram C Z is double of the triangle I C A, as they are upon the same base C I, and between the same parallels (XLI); and the parallelogram C E is double of the triangle B C F, as they are upon the same base C F, and between the same parallels (XLI); therefore the parallelograms C Z and C E, being double of the equal triangles I C A and B C F, are equal to one another. In the same manner it can be demonstrated, that A X and A E are equal, therefore the whole D A C F is equal to the sum of C Z and A X.
★★★ (203) Cor. 1.—Hence if the sides of a right angled triangle be given in numbers, its hypotenuse may be found; for let the squares of the sides be added together, and the square root of their sum will be the hypotenuse (187).
★★★ (204) Cor. 2.—If the hypotenuse and one side be given in numbers, the other side may be found; for let the square of the side be subtracted from that of the hypotenuse, and the remainder is equal to the square of the other side. The square root of this remainder will therefore be equal to the other side.
(205) Cor. 3.—Given any number of right lines, to find a line whose square is equal to the sum of their squares. Draw two lines A B and B C as right angles, and equal to the first two of the given lines, and draw A C. Draw C D equal to the third and perpendicular to A C, and draw A D. Draw D E equal to the fourth and perpendicular to A D, and draw A E, and so on. The square of the line A E will be equal to the sum of the squares of A B, B C, C D, &c., which are respectively equal to the given lines.
For the sum of the squares of A B and B C is equal to the square of A C. The sum of the squares of A C and C D, or the sum of the squares of A B, B C, C D is equal to the square of A D, and so on; the sum of the squares of all the lines is equal to the square of A E.
(206) Cor. 4.—To find a right line whose square is equal to the difference of the squares of two given right lines.
Through one extremity A of the lesser line A B draw an indefinite perpendicular A C; and from the other extremity B inflect on A C a line equal to the greater of the given lines (60); which is always possible, since the line so inflected is greater than B A, which is the shortest line which can be drawn from B to A C. The square of the intercept A D will be equal to the difference of the squares of B D and B A, or of the given lines.
(207) Cor. 5.—If a perpendicular (B D) be drawn from the vertex of a triangle to the base, the difference of the squares of the sides (A B and C B) is equal to the difference between the squares of the segments (A D and C D). For the square of A B is equal to the sum of the squares of A D and B D, and the square of C B is equal to the sum of the squares of C D and B D. The latter being taken from the former, the remainders, which are the difference of the squares of the sides A B and C B, and the difference of the squares of the segments A D and C D, are equal
(208) To understand this corollary perfectly, it is necessary to attend to the meaning of the term segments. When a line is cut at any point, the intercepts between the point of section and its extremities are called its segments. When the point of section lies between the extremities of the line it is said to be cut internally; but when, as sometimes happens, it is not the line itself but its production that is cut, and therefore the point of section lies beyond one of its extremities, it is said to be cut externally. By due attention to the definition of segments given above, it will be perceived that when a line is cut internally, the line is the sum of its own segments: but when cut externally, it is their difference.
The case of a perpendicular from the vertex on the base of a triangle offers an example of both species of section. If the perpendicular fall within the triangle, the base is cut internally by it; but if it fall outside, it is cut externally. In both cases the preceding corollary applies, and it is established by the same proof. the segments are in each case the intercepts A D and C D between the perpendicular and the extremities of the base.
(209) Cor. 6.—If a perpendicular be drawn from the vertex B to the base, the sums of the squares of the sides and alternate segments are equal.
For the sum of the squares of A B and B C is equal to the sum of the squares A B, B D and C D, since the square of B C is equal to the sum of the squares of B D and D C. For a similar reason, the sum of the squares of A B and B C is equal to the sum of the squares of A D, D B and B C. Hence the sum of the squares of A B, B D and D C is equal to that of A D, B D and B C. Taking the square of B D from both, the sum of the squares of A B and C D is equal to that of B C and A D.
Whether we consider the 47th proposition with reference to the peculiar and beautiful relation established in it, or to its innumerable uses in every department of mathematical science, or to its fertility in the consequences derivable from it, it must certainly be esteemed the most celebrated and important in the whole of the elements, if not in the whole range of mathematical science. It is by the influence of this proposition, and that which establishes the similitude of equiangular triangles (in the sixth book), that Geometry has been brought under the domininon of Algebra, and it is upon these same principles that the whole science of Trigonometry is founded.
The XXXIId and XLVIIth propositions are said to have been discovered by Pythagoras, and extraordinary accounts are given of his exultation upon his first perception of their truth. It is however supposed by some that Pythagoras acquired a knowledge of them in Egypt, and was the first to make them known in Greece.
Besides the demonstration in the Elements there are others by which this celebrated proposition is sometimes established, and which, in a principle of such importance, it may be gratifying to the student to know.
★★★ (210) 1° Having constructed squares on the sides A B, B C on opposite sides of them from the triangle, produce I H and F G to meet at L. Through A and C draw perpendiculars to the hypotenuse, and join K O.
In the triangles A F K and A B C, the angles F and B are equal, being both right, and F A K and B A C are equal, having a common complement K A B, and the sides F A and F B are equal. Hence A K and A C are equal, and in like manner C O and A C are equal. Hence A O is an equilateral parallelogram, and the angle at A being right, it is a square. The triangle L G B is, in every respect, equal to B C A, since B G is equal to B A, and L G is equal to B H or B C, and the angle at G is equal to the right angle B. Hence it is also equal in every respect to the triangle K F A Since, then, the angles G L B and F K A are equal, K A is parallel to B L, and therefore A L is a parallelogram. The square A G and the parallelogram A L are equal, being on the same base A B, and between the same parallels (XXXV); and for the same reason the parallelograms A L and K N are equal, A K being their common base. Therefore the square A G is equal to the parallelogram K N
In like manner the square C H is equal to the parallelogram O N, and therefore the squares A G and C H are together equal to A O.
★★★ (211) 2° Draw A G perpendicular and equal to A C, and produce B A, and draw G D perpendicular to it. In the same manner draw C H perpendicular and equal to C A, and produce B C and draw H F perpendicular to it. Produce F H and D G to meet in E, and draw G H.
The triangles G D A and H F C are equal in every respect to A B C (XXVI). Hence F C, G D and A B are equal, and also H F, D A and B C, and the angles in each triangle opposed to these sides are equal. Also, since G A and H C are equal to A C, and therefore to each other, and the angles at A and C are right, A H is a square (XXXIII). Since G H is equal to A C, and the angles at G and H are right, it follows that the triangle G E H is in all respects equal to A B C (XXVI), in the same manner as for the triangles G D A and H F C.
Through C and A draw the lines C K and A L parallel to B D and B F. Since C B and A I are equal and also C B and A D, it follows that A K is the square of B C, and in like manner that C L is the square of A B. The parallelograms B I and K L have bases and altitudes equal to those of the triangle A B C, and are therefore each equal to twice the triangle, and together equal to four times the triangle. Hence B I and K L are together equal to A B C, C F H, H E G and G D A together. Taking the former and the latter successively from the whole figure, the remainders are in the one case the squares D I and C L of the sides B C and B A, and in the latter the square A H of the hypotenuse. Therefore, &c.
(212) 3° On the hypotenuse A C construct the square A H, and draw G D and H E parallel to C B and A B, and produce these lines to meet in F, E and D. The triangles A B C, A D G, G E H and H F C are proved in every respect equal (XXVI). It is evident, that the angles D, E, F, B are all right. But also since D G and A B are equal, and also G E and A D, taking the latter from the former D E and D B remain equal. Hence B E is a square on the difference D B of the sides; and therefore the square of A C is divided into four triangles, in all respects equal to A B C and the square B E of the difference of the sides.
Now let squares B G and B I be constructed on the sides, and taking A E on the greater side equal to B C the less, and draw E H parallel to B C, and produce G C to K. Draw G E and A H
The part B E is the difference of the sides A B and B C. And since B F is equal to A B, F C is also the difference of the sides, wherefore F L is the square of this difference. Also since A E and B D are equal A B and D E are equal, therefore the parallelogram D L is double the triangle A B C. The sides and angles of the parallelogram A H are equal respectively to those of D L, and therefore these two parallelograms together are equal to four times the triangle A B C. Hence the squares A F and B G may be divided into four triangles G D E, G L E, A E H and A I H in all respects equal to the triangle A B C, and the square C H of the difference of the sides. But by the former construction the square of the hypotenuse was shown to be divisible into the same parts. Therefore, &c.
The peculiarity of this proof is, that it shows that the squares of the sides may be so dissected that they may be laid upon the square of the hypotenuse so as exactly to cover it, and vice versa, that the square of the hypotenuse may be so dissected as to exactly cover the squares of the sides.
(213) The forty-seventh proposition is included as a case of the following more general one taken from the mathematical collections of Pappus, an eminent Greek Geometer of the fourth century.
In any triangle (A B C) parallelograms A E and C G being described on the sides, and their sides D E and F G being produced to meet at H, and H B I being drawn, the parallelogram on A C whose sides are equal and parallel to B H is equal to A E and C G together.
For draw A K and C L parallel to B H, to meet D H and F H in K and L. Since A H is a parallelogram, A K is equal to B H, and for a similar reason C L is equal to B H. Hence C L and A K are equal and parallel, and therefore (XXXIII) A L is a parallelogram. The parallelograms A E and A H are equal, being on the same base A B, and between the same parallels, and also A H and K I whose common base is A K. Hence the parallelograms A E and K I are equal. In like manner the parallelograms C G and L I are equal, and therefore A E and C G are together equal to A L.
This proof is applied to the forty-seventh in (210).
In any triangle (A B C) squares being constructed on the sides (A B and B C) and on the base; and perpendiculars (A D F and C E G) being drawn from the extremities of the base to the sides, the parallelograms A G and C F formed by the segments C D, A E, with the sides of the squares, will be together equal to the square of the base A C.
For draw A H and B I; and also B K perpendicular to A C.
The parallelograms K C and C F are proved equal, exactly as C E and C Z are proved equal in the demonstration of the XLVIIth. And in like manner it follows, that A K and A G are equal, and therefore the square on A C is equal to the parallelograms A G and C F together.
If the triangle be right angled at B, the lines G E and D F will coincide with the sides of the squares. and the proposition will become the XLVIIth.
(215) If B be acute the perpendiculars A D and C E will fall within the triangle, and the parallelograms A G and C F are less than the squares of the sides; but if B be obtuse the perpendiculars fall outside the triangle, and the parallelograms A G and C F are greater than the squares of the sides.
Hence the forty-seventh proposition may be extended thus:
The square of the base of a triangle is less than, equal to, or greater than the sum of the squares of the sides, according as the vertical angle is less than, equal to, or greater than a right angle..
Book I: Euclid, Book I (ed. Dionysius Lardner, 11th Edition, 1855)
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