FE D Complete the parallelogram ABFC; C then the parallelogram ABFC is equivalent to the parallelogram ABDE, because they have the same base and the same altitude (Prop. I.). But the triangle ABC is half of the parallelogram ABFC (Prop. XXIX., Cor., B. I.); wherefore the triangle ABC is also half of the parallelogram ABDE. Therefore, every triangle, &c. . B Cor. 1. Every triangle is half of the rectangle which has the same base and altitude. Cor. 2. Triangles which have equal bases and equal alti tudes are equivalent. PROPOSITION III. THEOREM. Two rectangles of the same altitude, are to each other as their bases. Let ABCD, AEFD be two rec- D tangles which have the common altitude AD; they are to each other as their bases AB, AE. A E B Case first. When the bases are in the ratio of two whole numbers, for example, as 7 to 4. If AB be divided into seven equal parts, AE will contain four of those parts. At each point of division, erect a perpendicular to the base; seven partial rectangles will thus be formed, all equal to each other, since they have equal bases and altitudes (Prop. I.). The rectangle ABCD will contain seven partial rectangles, while AEFD will contain four; therefore the rectangle ABCD is to the rectangle AEFD as 7 to 4, or as AB to AE. The same reasoning is applicable to any other ratio than that of 7 to 4; therefore, whenever the ratio of the bases can be expressed in whole numbers, we shall have ABCD: AEFD::AB: AE. Case second. When the ratio of the bases can not be expressed in whole numbers, it is still true that ABCD: AEFD::AB: AE. D For, if this proportion is not true, the first three terms remaining the same, the fourth must be greater or less than AE. Suppose it to be greater, and that we have ABCD: AEFD ::AB: AG. Conceive the line AB to be divided into A FI C EHG B D FI C equal parts, each less than EG; there will ABCD: AHID :: AB: AH. But, by hypothesis, we have ABCD: AEFD::AB: AG. EHG B In these two proportions the antecedents are equal; therefore the consequents are proportional (Prop. IV., Cor., B. II.), and we have AHID AEFD :: AH: AG. But AG is greater than AH; therefore the rectangle AEFD is greater than AHID (Def. 2, B. II.); that is, a part is greater than the whole, which is absurd. Therefore ABCD can not be to AEFD as AB to a line greater than AE. In the same manner, it may be shown that the fourth term of the proportion can not be less than AE; hence it must be AE, and we have the proportion ABCD: AEFD::AB: AE. Therefore, two rectangles, &c. PROPOSITION IV. THEOREM. Any two rectangles are to each other as the products of their bases by their altitudes. Let ABCD, AEGF be two rectangles; the ratio of the rectangle ABCD to the rectangle AEGF, is the same with the ratio of the product of AB by AD, to the product of AE by AF; that is, E H D C A B ABCD: AEGF:: AB× AD: AEX AF. Having placed the two rectangles so that the angles at A are vertical, produce the sides GE, CD till they meet in H. The two rectangles ABCĎ, AEHD have the same altitude AD; they are, therefore, as their bases AB, AE (Prop. III.). So, also, the rectangles AEHD, AEGF, having the same altitude AE, are to each other as their bases AD, AF two proportions : ABCD AEHD:: AB G F Thus, we have the AE, AF. Hence (Prop. XI., Cor., B. II.), ABCD: AEGF:: ABAD: AEXAF. Scholium. Hence we may take as the measure of a rectangle the product of its base by its altitude; provided we understand by it the product of two numbers, one of which is the number of linear units contained in the base, and the other the number of linear units contained in the altitude. PROPOSITION V. THEOREM. The area of a parallelogram is equal to the product of its base by its altitude. E C Let ABCD be a parallelogram, AF its F D altitude, and AB its base; then is its surface measured by the product of AB by AF. For, upon the base AB, construct a rectangle having the altitude AF; the parallelogram ABCD is equivalent to the rectangle ABEF (Prop. I., Cor.). But the rectangle ABEF is measured by AB × AF (Prop. IV., Schol.); therefore the area of the parallelogram ABCD is equal to ABX AF. A B Cor. Parallelograms of the same base are to each other as their altitudes, and parallelograms of the same altitude are to each other as their bases; for magnitudes have the same ratio that their equimultiples have (Prop. VIII., B. II.). PROPOSITION VI. THEOREM. The area of a triangle is equal to half the product of its base by its altitude. A E Let ABC be any triangle, BC its base, and AD its altitude; the area of the triangle ABC is measured by half the product of BC by AD. For, complete the parallelogram ABCE. The triangle ABC is half of the parallelogram ABCE (Prop. II.); but the area of the parallelogram is equal to BCXAD (Prop. V.); hence the area of the triangle is equal to one half of the product of BC by AD. Therefore, the area of a triangle, &c. B D C Cor. 1. Triangles of the same altitude are to each other as their bases, and triangles of the same base are to each other as their altitudes. Cor. 2 Equivalent triangles, whose bases are equal, have equal altitudes; and equivalent triangles, whose altitudes are equal, have equal bases. PROPOSITION VII. THEOREM. The area of a trapezoid is equal to half the product of its altitude by the sum of its parallel sides. K D E H G B Let ABCD be a trapezoid, DE its altitude, AB and CD its parallel sides; its area is measured by half the product of DE, by the sum of its sides AB, CD. Bisect BC in F, and through F draw GH parallel to AD, and produce DC to A H. In the two triangles BFG, CFH, the side BF is equal to CF by construction, the vertical angles BFG, CFH are equal (Prop. V., B. I.), and the angle FCH is equal to the alternate angle FBG, because CH and BG are parallel (Prop. XXIII., B. I.); therefore the triangle CFH is equal to the triangle BFG. Now, if from the whole figure, ABFHD, we take away the triangle CFH, there will remain the trapezoid ABCD; and if from the same figure, ABFHD, we take away the equal triangle BFG, there will remain the parallelogram AGHD. Therefore the trapezoid ABCD is equivalent to the parallelogram AGHD, and is measured by the product of AG by DE. Also, because AG is equal to DH, and BG to CH, therefore the sum of AB and CD is equal to the sum of AG and DH, or twice AG. Hence AG is equal to half the sum of the parallel sides AB, CD; therefore the area of the trapezoid ABCD is equal to half the product of the altitude DE by the sum of the bases AB, CD. Cor. If through the point F, the middle of BC, we draw FK parallel to the base AB, the point K will also be the middle of AD. For the figure AKFG is a parallelogram, as also DKFH, the opposite sides being parallel. Therefore AK is equal to FG, and DK to HF. But FG is equal to FH, since the triangles BFG, CFH are equal; therefore AK is equal to DK. Now, since KF is equal to AG, the area of the trapezoid is equal to DEXKF. Hence the area of a trapezoid is equal to its altitude, multiplied by the line which joins the middle points of the sides which are not parallel. PROPOSITION VIII. THEOREM. If a straight line is divided into any two parts, the square of the whole line is equivalent to the squares of the two parts, together with twice the rectangle contained by the parts. Let the straight line AB be divided into any two parts in C; the square on AB is equivalent to the squares on AC CB, together with twice the rectangle contained by AC, CB; that is, AB3, or (AC+CB)2=AC2+CB2+2AC × CB. Upon AB describe the square ABDE; E take AF equal to AC, through F draw FG parallel to AB, and through C draw CH par- F allel to AE. Ꭺ . H D I G C B The square ABDE is divided into four parts: the first, ACIF, is the square on AC, since AF was taken equal to AC. The second part, IGDH, is the square on CB; for, because AB is equal to AE, and AC to AF, therefore BC is equal to EF (Axiom 3, B. I.). But, because BCIG is a parallelogram, GI is equal to BC; and because DEFG is a parallelogramı, DG is equal to EF (Prop. XXIX., B. I.); therefore HIGD is equal to a square described on BC. If these two parts are taken from the entire square, there will remain the two rectangles BCIG, EFIH, each of which is measured by ACX CB; therefore the whole square on AB is equivalent to the squares on AC and CB, together with twice the rectangle of ACXCB. Therefore, if a straight line, &c. Cor. The square of any line is equivalent to four times the square of half that line. For, if AC is equal to CB, the four figures AI, CG, FH, ID become equal squares. Scholium. thus: This proposition is expressed algebraically (a+b)2=a2+2ab+b2. PROPOSITION IX. THEOREM. The square described on the difference of two lines, is equivalent to the sum of the squares of the lines, diminished by twice the rectangle contained by the lines. Let AB, BC be any two lines, and AC their difference; the square described on AC is equivalent to the sum of the |