Proposition XL. Theorem.
|(173)||Equal triangles (B A C and E D F) on equal bases and on the same side, are between the same parallels.|
For if the right line A D which joins the vertices of the two triangles be not parallel to B F, draw through the point A the right line A G parallel to B F, cutting a side D E of the triangle E D F, or the side produced in a point G different from the vertex, and join F G.
Because the right line A G is parallel to B F, and B C and E F are equal, the triangle G E F is equal to B A C (XXXVIII); but E D F is also equal to B A C (hyp.), therefore E G F and E D F are equal; a part equal to the whole, which is absurd. Therefore A G is not parallel to B F, and in the same manner it can be demonstrated, that no other line except A D is parallel to B F, therefore A D is parallel to B F.
From this and the preceding propositions may be deduced the following corollaries.
(174) Cor. 1.—Perpendiculars being drawn through the extremities of the base of a given parallelogram or triangle, and produced to meet the opposite side of the parallelogram or a parallel to the base of the triangle through its vertex, will include a right angled parallelogram which shall be equal to the given prallelogram; and if the diagonal of this right angled parallelogram be drawn, it will cut off a right angled triangle having the same base with the given triangle and equal to it. Hence any parallelogram or triangle is equal to a right angled parallelogram or triangle having an equal base and altitude.
(175) Cor. 2.—Parallelograms and triangles whose bases and altitudes are respectively equal are equal in area.
(176) Cor. 3.—Equal parallelograms and triangles on equal bases have equal altitudes.
(177) Cor. 4.—Equal parallelograms and triangles in equal altitudes have equal bases.
(178) Cor. 5.—If two parallelograms or triangles have equal altitudes, and the base of one be double the base of the other, the area of one will be also double the area of the other. Also if they have equal bases and the altitude of one be double the altitude of the other, the area of the one will be double the area of the other.
(179) Cor. 6.—The line joining the points of bisection fo the sides of a triangle is parallel to the base.
For if lines be drawn from the extremities of the bse to the points of bisection they will each bisect the area (170) of the triangle; therefore the triangles having the base of the given triangle as a common base and their vertices at the middle points of the sides, are equal, and therefore between the same parallel.
(180) Cor. 7.—A parallel to the base of a triangle through the point of bisection of one side will bisect the other side.
For by the last Cor. the line joining the points of bisection of the sides is parallel to the base, and two parallels to the same line cannot pass through the same point.
(181) Cor. 8.—The lines which join the middle points D E F of the three sides of a triangle divide it into four triangles which are equal in every respect.
(182) Cor. 9.—The line joining the points of bisection of each pair of sides is equal to half of the third side.
★★★ (183) Cor. 10.—If two conterminous sides of a parallelogram be divided each into any number of equal parts, and through the several points of division of each side parallels be drawn to the other side, the whole parallelogram will be divided into a number of equal parallelograms, and this number is found by multiplying the number of parts in one side by the number of parts in the other. This is evident from considering, that by the parallels through the points of division of one side the whole parallelogram is resolved into as many equal parallelograms as there are parts in the side through the points of which the parallels are drawn; and the parallels through the points of division of the other side resolve each of these component parallelograms into as many equal parallelograms as there are parts in the other side. Thus the total number of parallelograms into which the entire is divided, is the product of the number of parts in each side.
★★★ (184) Cor. 11.—The square on a line is four times the square of its half.
★★★ (185) Cor. 12.—If the sides of a right angled parallelogram be divided into any number of equal parts, and such that the parts of one side shall have the same magnitude as those of the other, the whole parallelogram will be equal to the square of one of the parts into which the sides are divided, multiplied by the product of the number of parts in each side. Thus, if the base of the parallelogram be six feet and the altitude be eight feet, the area will be one square foot multiplied by the product of six and eight or forty-eight square feet. In this sense the area of such a parallelogram is said to be found by multiplying its base by its altitude.
★★★ (186) Cor. 13.—Also, since the area of any parallelogram is equal to that of a right-angled parallelogram having the same base and altitude, and that of a triangle is equal to half that area, it follwos that the area of a parallelogram is the product of its base and its altitude, and that of a triangle is equal to half that product.
The phrase ‘the product of two lines,’ or ‘multiplying one line by another, is only an abridged manner of expressing the multiplication of the number of parts in one of the lines by the number of parts in the other. Multiplication is an operation which can only be effected, properly speaking, by a number and not by a line
★★★ (187) Cor. 14.—The area of a square is found numerically by multiplying the number of equal parts in the side of the square by itself. Thus a square whose side is twelve inches contains in its area 144 square inches. Hence, in arithmetic, when a number is multiplied by itself the product is called its square. Thus 9, 16, 25, &c. are the squares of 3, 4, 5, &c.; and 3, 4, 5,&c. are called the square roots of the numbers 9, 16, 25, &c. Thus square and square root are correlative terms.
★★★ (188) Cor. 15.—If the four sides of a quadrilateral A B C D be bisected, and the middle points E F H G of each pair of conterminous sides joined by right lines, those joining lines will form a parallelogram E F H G whose area is equal to half that of the quadrilateral.
Draw C A and B D. The lines E F and G H are parallel to C A (179), and equal to half of C A (182). Therefore E F and G H are equal and parallel, and therefore (XXIII) E F H G is a parallelogram. But E B F is one-fourth of C B A and G H D one fourth of C D A (181), and therefore E B F and G D H are together one-fourth of the whole figure. In like manner E C G and F A H are together one-fourth of the whole, and therefore F B E, E C G, G D H, and H A F are together one-half of the whole figure, and therefore E F H G is equal to half the figure.
★★★ (189) Cor. 16.—A trapezium is equal to a parallelogram in the same altitude, and whose base is half the sum of the parallel bases.
Let C D be bisected at H, and through H draw G F parallel to A B.
Since C G and F D are parallel, the angles G C H and G are respectively equal to D, and H F D (XXIX) and C H is equal to H D, therefore (XXVI) C G is equal to F D, and the triangle C H G to the triangle D H F. Therefore A F and B G are together equal to A D and B C, and the parallelogram A G to the trapezium A C; and since A F and B G are equal, A F is half the sum of A D and B C.
Book I: Euclid, Book I (ed. Dionysius Lardner, 11th Edition, 1855)
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