289. If from the square of the sum of two lines we take away the sum of their squares, we shall have left twice their rectangle. 290. Scholium. By hypothesis we have AB = AP + PB. Substituting this in the conclusion, we have (AP + PB)2 = AP2 + 2AP . PB + PB2, a well-known algebraic expression. The geometric construction serves to exhibit to the eye the different parts of which this algebraic expression is made up. THEOREM III. 291. The square upon the difference of two lines is equal to the sum of the squares upon the lines, diminished by twice the rectangle contained by them. Hypothesis. AB, AC, two lines of which AC is the longer; BC, their difference. Gr On BC erect the square BCEF. D D. Then 1. The whole area AKLBCHG = AB2 + AC2. 2. Because EB = BC, and BL = AB, we have EL = AB + BC = AC. Therefore E H 4. If from the whole area (1) we take away the areas (2) and (3), we have left the square BCEF; that is, BC2. Therefore BC = AB* +AC – 2AB.AC. Q.E.D. 292. Scholium. Since BC = AC — AB, we have (AC — AB)2 = AC-2AB. AC+AB2, the algebraic formula for the square of the difference of two numbers. THEOREM IV. 293. The difference of the squares described on two lines is equal to the rectangle contained by the sum and difference of the lines. Hypothesis. AB, AC, two straight lines of which AC is the greater, and each of which is to have a square described upon it. Conclusion. Er D 1. In the same way as in the last theorem it may be shown that EH and GC are rectangles, and AG a square. 2. Because AE AC (by construction), and AF AB, FE is equal to AC AB, and ED is, by construction, equal to AC. Therefore = = 3. Because FH and AC are parallel, CH= AF AB, while BC is, by construction, equal to AC- AB. Therefore Rectangle GC AB (AC-AB). 4. The sum of the rectangles AC (AC-AB) and AB (AC — AB) = rectangle (AC+ AB) (AC — AB). (§ 287) 5. The difference between the squares on the lines AB and AC is made up of the sum of these rectangles. Therefore ACAB (AC+ AB) (AC-AB). Q.E.D. = 294. Scholium. Expressing the areas of the squares and rectangles in algebraic language, this theorem gives a2 — b2 = (a + b) (a - b). CHAPTER II. AREAS OF PLANE FIGURES. THEOREM V. 295. The area of a parallelogram is equal to that of the rectangle contained by its base and its altitude. E F C E C F Hypothesis. ABCD, any parallelogram of which the side AB is taken as the base; AE, the altitude of the parallelogram. Conclusion. Area ABCD = rectangle AB. AE. Proof. From A and B draw perpendiculars to the base AB, meeting CD produced in E and F. Then 1. Because ABCD and ABEF are both parallelograms, EF AB, and CD = AB. Therefore (§ 127) 2. If from the line ED we take away EF, FD remains; and if we take away CD, EC remains. Because the parts taken away are equal (1), FDEC. 3. Comparing with the last two equations of (1), it is seen that the triangles BFD and AEC have the three sides of the one equal to the three sides of the other. Therefore Triangle AEC = triangle BFD. 4. From the trapezoid ABED take away the triangle AEC, and there is left the parallelogram ABCD. From the same trapezoid take away the equal triangle BFD, and there is left the rectangle ABEF. Because the triangles are equal, Rectangle ABEF = parallelogram ABCD. Therefore Area ABCD rectangle AB. AE. Q.E.D. = 296. Corollary 1. All parallelograms upon the same base and between the same parallels are equal in area, because they are all equal to the same rectangle. 297. Cor. 2. Parallelograms having equal bases and equal altitudes are equal in area. 298. Cor. 3. Of two parallelograms having equal bases, that has the greater area which has the greater altitude. Of parallelograms having equal altitudes, that has the greater area which has the greater base. THEOREM VI. 299. The area of a triangle is equal to half the area of the parallelogram formed from any two of its sides, having an angle equal to that between those sides. Hypothesis. ABC, any triangle; PQRS, a parallelogram in which Proof. Draw the diagonal PS. area PQRS. Then 1. Because of the equations supposed in the hypothesis, the triangles PQS and ABC have two sides and the included angle of the one equal to two sides and the included angle of he other. Therefore the triangles are identically equal, and Area ABC area PQS. 2. In the same way is shown Area ABC = area PRS. (§ 108) 3. The sum of the areas PQS and PRS makes up the whole area of the parallelogram PQRS. Therefore, comparing (1) and (2), Area ABC area PQRS. Q.E.D. 300. Corollary. A diagonal of a parallelogram divides it into two triangles of equal area. THEOREM VII. 301. The area of a triangle is one half the area of the rectangle contained by its base and its altitude. Hypothesis. ABC, a triangle having the base AB and the altitude CD. 1. ABCG is a parallelogram having the base AB and the altitude CD. Therefore Area ABCG = rect. AB. CD. (§ 295) 2. Because AB and CD are each sides of this parallelo gram, Area ABC = area ABGC. (§ 299) |