E W (21.), and EBCH is a parallelogram. In the second place, therefore, let the But the parallelogram E B CH is equal triangles ABC, D B C, standing upon to ABCD, because it is upon the same the same base B C, or upon equal bases base BC, and between the same paral- BC, B C, in the same straight line, and lels (24.): and for the like reason EBCH towards the same parts, be equal to one is equal to EFGH. Therefore (ax. 1.) another; and let A D be joined: AD ABCD is equal to EFGH. shall be parallel to BC. Therefore, &c. For, if not, let A E (14. Cor. 1.) be Cor. The squares of equal straight parallel to BC; and let it meet D B in lines are equal to one another: and E. Join E C. Then, by the first part conversely. of the proposition, because ABC, PROP. 26. (Euc. i. 41.) E B C, are upon the same base, or If a parallelogram and a triangle be upon equal bases, and between the same parallels, the triangle E B C is upon the same base and between the equal to A B C, that is, to DBC; the same parallels, the parallelogram shall less to the greater, which is imposbe double of the triangle. sible. Therefore A E is not parallel to Let the parallelo BC; and in the same manner it may be gram ABCD and the shown that no other straight line which triangle E B C be upon passes through A, except A D only, can the same base B C, and be parallel to BC; that is, (14. Cor. 1.) between the same pa AD is paralle) to B C. rallels AD, BC. The Therefore, &c. parallelogram ABCD shall be double Cor. 1. It is evident that the second of the triangle E BC. (or converse) part of the proposition apComplete the parallelogram E B CF. plies equally to parallelograms, as to (14. Cor. 1.) Then the parallelogram triangles. EBC F is double of the triangle EBC, Cor. 2. If a quadrilateral be bisected because it is bisected by the diagonal E C by each of its diagonals, it must be a (22. Cor.); and ABCD is equal to E B parallelogram : for the two triangles CF, because it is upon the same base, ABC, DBC (see the figure of Prop. and between the same parallels (24.). 21.) which stand upon any one of its Therefore, the parallelogram ABCD sides BC for a base, and which have their is also double of the triangle E B C. vertices in the side opposite, being equal, Therefore, &c. each of them, to half the quadrilateral, Cor. Every triangle is equal to the are equal to one another; and therefore half of a rectangle of the same base and A D is parallel to BC; and, for a like altitude. reason, A B is parallel to DC. PROP. 27. (Euc. i, 37, 38, 39, & 40.) PROP. 28. Triangles upon the same base, or upon equal bases, and between the same A trapezoid is equal to the half of a parallels, are equal to one another : and rectangle having the same altitude, and conversely, equal triangles, upon the a base equal to the sum of its parallel same base, or upon equal bases in the sides. same straight line, and towards the same Let ABCD be a traparts, are between the same parallels. pezoid, having the side The first part of the AD parallel to the side proposition is manifest; BC. The trapezoid for the triangles are ABCD shall be equal to the halves of parallelo the half of a rectangle grams (26.) upon the B having the same altitude, and a base same base, or upon equal bases, and equal to the sum of AD, BC. between the same parallels; and because Join AC; through D draw D E pathese parallelograms are equal to one rallel to AC, and let it meet the base another (24. or 25.), the triangles, which B C produced in E, and join A E. are their halves, are also equal (ax. 5.). Then, because D E is parallel to AC, the triangle A C D is (27.) equal to ACE. * To complete the parallelogram, in this case, it is only requisite that C F should be drawn through Therefore, the triangles ABC, ACD, in F. The word complete, indeed, almost explains AB C, ÁC E, together (ax. 2.); or, the the point e parallel to B E, to meet A D produced together, are equal to the triangles itself: in future constructions it will be introduced without further notice. trapezoid ABCD is equal to the tri С a D B square, &c. E M angle A BE. But the base of the tri- under those two straight lines shall be angle A B E is equal to the sum of AD, contained as often in the given rectangle, B!', because (22.) C E is equal to AD; as is denoted by the product of the two and (26. Cor.) every triangle is equal to numbers which denote how often the lines the half of a rectangle of the same themselves are contained in the two base and altitude. Therefore, the trape- sides. zoid ABCD is equal to the half of a Cor. 2. The square of twice M is rectangle of the same altitude, and upon equal to 4 times M square, because it is a base which is equal to the sum of AD, a rectangle, in which each of the sides BC. contains M twice. In like manner, the Therefore, &c. square of 3 times M is equal to 9 times PROP. 29. M square-of 4 times M to 16 times M square-of 5 times M to 25 times M If the adjoining sides of a rectangle contain, each of them, the same straight Cor. 3. The square of 5, or 25, is equal line, a certain number of times exactly, to the sum of 16 and 9. Consequently the rectangle shall contain the square of the square of 5 times M is equal to the that straight line, as often as is denoted square of 4 times M, together with the by the product of the two numbers, square of 3 times M. which denote how often the line itself is contained in the two sides. Scholium. Let ABD C be a From the theorems of this Section rectangle, and let its rules are easily deduced for the menadjacent sides A B, suration of rectilineal figures. For AC, contain each of every rectilineal figure may be divided them the straight into triangles; and every triangle, being line M a certain equal (26. Cor.) to half the rectangle number of times ex under its base and altitude, contains actly, viz., AB 6 half as many square units as is denoted times, and AC4 by the product of the numbers which times: the rectangle ABDC shall con- express how often the corresponding tain the square of M, 6 X 4, or 24 times. linear unit is contained in its base and Divide A B, A C, each of them, into in its altitude. Let this linear unit be, parts equal to M; and, through the divi- for example, a foot; and let it be resion-points of each, draw straight lines quired to find how many square feet parallel to the other, thereby dividing there are in a triangle whose altitude is the rectangle into six upright rows of 10 feet, and its base 9 feet. The rectfour parallelograms each, that is, upon angle, of which these are the sides, the whole, into twenty-four parallelo- contains 10 x 9, or 90 square feet (by grams. Now these parallelograms are Prop. 29.); and, therefore, the triangle all of them rectangular, because their contains 45 square feet. containing sides are parallel to AB, AC, Hence, a rectangle is sometimes said the sides of the right angle A (18.). They to be equal to the product of its base are also equilateral: for any one of them, and altitude, a triangle to half the proas E, has its upright sides each of them duct of its base and altitude, and the (22,) equal to a division of A C, that is, like; expressions which must be underto M; and its other two sides each of stood as above, the words rectangle, &c. them equal to a division of A B, that is, base, &c. being briefly put for the numto M. Therefore, they are squares ber of square units in the rectangle, fc. (def. 20.), equal, each of them, to the the number of linear units in the base, square of M. And they are twenty-four foc. By the length of a line is comin number. Therefore, the square of M monly understood the number of linear is contained twenty-four times in the units which it contains; and the term rectangle A B D C. superficial area, or area, is similarly The same may be said, if, instead of 6 applied to denote the number of square and 4, any other two numbers be taken. units in a surface. With regard to the Therefore, &c. measuring unit, a less or a greater is Cor. 1. In like manner, it may be convenient, according to the subject shown, that, if there be two straight lines, of measurement : a glazier measuring one of which is contained an exact num- his glass by square inches, a carpenter ber of times in one side of a rectangle, his planks by square feet, a proprietor and the other an exact number of times his land by acres, and a geographer the in the side adjoining to it; the rectangle extent of countries by square miles. A E B IH H K EB. Section 5. Rectangles under the parts is equal to BD, that is to CB; and the rectangle AC, CB, because (22.) CF of divided lines. figure CD is equal to the square of PROP. 30. (Euc. ii. 1.) CB; and the figure AD to the rectanIf there be two straight lines, one of gle A B, B.C. But A D is equal to A F, which is divided into any number of together with CD. Therefore, the recparts, the rectangle contained by the tangle A B, B C, is equal to the rectangle two lines shall be equal to the sum of AC, C B, together with the square of the rectangles contained by the undi BC.* vided line, and the several parts of the Therefore, &c. divided line. Let A B and C be two straight lines, PROP. 32. (Euc. ii. 4.) of which A B is divided into the parts The square of the sum of two lines is AD, DE, EB: the rectangle under C greater than the sum of their squares, and A B shall be equal by twice their rectangle. to the sum of the rec Let the straight line A B be the sum tangles under C and of the two straight lines AD, C and D E, C and AC, CB: the square of A B shall be greater than Draw the straight line A F at right the squares of AC, CB, angles to A B, and equal to C (post. 5.): by twice the rectangle complete the rectangle AG, and through AC, C B. the points D and É draw the straight Because the straight line AB is lines DH and EK parallel to ÅF divided into two parts in the point C, (14. Cor.). Then, because D H and EK (30. Cor.) the square of A B is equal to are each of them (22.) equal to AF, the sum of the rectangles under A B, that is, to C, the rectangles AH, DK, AC, and A B, BC. But the rectangle EG are equal to the rectangles under under A B, A C (31.) is equal to the C and A Ď, C and DE, C and EB. rectangle under A C, C B together with But these rectangles make up the whole the square of A C; and, in like manner, rectangle AG, which is equal to the the rectangle under A B, B C is equal rectangle under C and A B. Therefore, to the rectangle under A C, C B together the rectangle under C and A B is equal with the square of C B. Therefore, the to the rectangles under C and AD, C square of A B is equal to twice the rectand D E, C and E B. angle AC, CB, together with the Therefore, &c. squares of AC, CB; or, which is the Cor. (Euc. ii. 2.) If a straight line same thing, the square of A B is greater be divided into any two parts, the rect- than the squares of AC, C B, by twice angles contained by the whole line and the rectangle A C, C B. each of the parts, shall be together equal Therefore, &c. to the square of the whole line. The figure shews in what manner the square of AB may be divided into two PROP. 31. (Euc. ii. 3.) squares equal to those of A C, C B, and If a straight line be divided into any two rectangles, each equal to the recttwo parts, the rectangle contained by angle A C, C B. the whole line and one of the parts, PROP. 33. (Euc. č. 7.) shall be equal to the rectangle contained by the two parts, together with the The square of the difference of 'two lines is less than the sum of their squares square of the aforesaid part. Let the straight line A B be divided by twice their rectangle. into any two parts AC, CB: the rec- Let the straight line, 古 tangle under AB, BC A B, be the difference shall be equal to the rec of the two straight lines, AC, CB: the tangle under AC, CB, square of A B shall be less than the together with the square squares of A C, C B, by twice the rectof BC. angle A C, C B. Draw the straight line B D at right * This proposition, being an obvious consequence angles to A B (post. 5.), and equal to BC: of the preceding, might have been added to it as a complete the rectangle ABDE and second corollary: it is of so great importance, how ever, that it seemed preferable to force it upon the through C draw C F parallel to B D. attention of the student, by placing it among the Then, the figure A F is equal to the propositions. A A c E Because AC is the sum of A B and together with twice the rectangle AB, BC (32.), the square of A C is equal to BC. Therefore (ax. 1.) the difference the squares of AB, BC, together with of the squares of A C, A B, is equal to twice the rectangle, A B, BC: therefore the rectangle CF, that is, to the rectthe square of A B is less than the square angle under CD, C B. of AC by the square of B C, together Therefore, &c. with twice the rectangle AB, BC. Therefore, the same square of A B is less PROP. 33. (Euc. ii. 9. and 10.) than the squares of A C, B C, by twice the square of B C, together with twice difference of two lines, are together The squares of the sum, and of the the rectangle A B, BC. But twice the double of the squares of the two lines. square of BC, together with twice the rectangle A B, BC, is (32.) equal to lines is greater than the sum of their For (32.) the square of the sum of two twice the rectangle A C, C B. fore, the square of A B is less than the squares, by twice their rectangle, and the squares of A C, B C, by twice the rect- square of their difference is (33.) as much less than the sum of their squares. angle AC, CB. Therefore (ax. 9.), the square of the Therefore, &c. Cor. (Euc. ii . 8.) The square of the sum, together with the square of the sum of two lines is greater than the difference, is equal to twice the sum of their squares. square of their difference by four times their rectangle : for the former square is Therefore, &c. greater than the sum of their squares by Scholium. twice their rectangle (32.), and the latter The theorems of this section admit of square is less than the sum of their being enunciated more briefly and persquares by twice their rectangle. spicuously by the use of certain convenPROP. 34. (Euc. ii. 5.) tional signs =, +, -, X,( ), &c. bor rowed from Algebra. The difference of the squares of two That A is equal to B, is thus denoted : lines is equal to the rectangle under A = B, which is read “A is equal to B." their sum and difference. The sum of A and B thus: A +B, Let A B, AC, which is read “ A plus B." be any two straight The excess of A above B:A B....... lines, and let BA “ A minus B." be produced to D, Twice A, four times A, &C.........2 A, so that A D may be equal to A B, and 4 A, &c. therefore C D equal to the sum, and The rectangle under A and B...AXB, C B to the difference of A B, AC: the or A B.......“ A, B." difference of the squares of A B, AC, The square of A........A A, or A2......... shall be equal to the rectangle under A square ; " and, CD, CB. (A + B - C) or A + B-C signifies Draw the straight line C H perpen- that A + B-C (that is, the excess of dicular to C D (post. 5.), and equal to the sum of A and B above C) is to be CB; complete the rectangle CHFD; be taken as a single quantity. and, through the points A and B, draw AG and B E parallel to CH. The theorems of this section may, Then, because BE is (22.) equal to therefore, be more briefly expressed as CH, that is, to B C, the rectangle A E follows:is equal to the rectangle A B, BC; Prop. 30. Ax (B+C+D) = AB+ and, because AD is equal to A B, the AC+AD. rectangle D G is equal (25.) to the rect- Prop. 31. (A+B) x B=AB+B%. angle AE; and CE is the square of Prop. 32. (A+B)2=A2+B2+2 A B. CB: therefore, the whole rectangle Prop. 33. (A-B)2=A2+B2_2 A B. CF is equal to the square of BC, to- Prop. 33. Cor. · (A+B) -(A-B)= gether with twice the rectangle A B, BC. 4 A B. But, because the square of A C is equal Prop. 34. (A+B) X(A-B)=A2-B2. (32.) to the squares of A B, BC, together Prop. 35. (A+B)2+(A–B) = 2 (A2 with twice the rectangle A B, BC; if the +B2). square of A B be taken from each side, In this borrowing of its notation, may (ax. 3.) the difference of the squares of be seen the first glimmerings of the apÀ C, A B, is equal to the square of B C, plication of Algebra to Geometry. It D B C G E H H requires but a very slight acquaintance is equal to the rectangle BK (ax. 1.). with the former science, to perceive at And, in like manner, it may be shewn once, that, when the lines contain each that the square of B C is equal to the of them some common part Ma cer- rectangle KC. Therefore, the squares tain number of times exactly, these of AB, AC, together, are equal to the theorems are but so many examples of rectangles B K, KC together, that is, to its rules of addition, subtraction, and the square of B C. multiplication. The perspicuity alone, Next, let the square of B C be equal however, which is displayed in the above to the sum of the squares of BA, AC; expressions, will enable the uninitiated the angle B A C shall be a right angle. reader to form some notion of the ad- For, let L M N be another triangle having vantages resulting from a more intimate its sides, L M, LN, equal to the sides AB, union of the two sciences. AC, respectively, and the angle at L a right angle. Then, by the former part Section 6. Relations of the sides of of the proposition, the square of MN is Triangles. equal to the squares of LM, LN, that is, to the squares of A B, AC, or to the PROP. 36. (Euc. i. 47. and 48.) square of B C, and M N is equal to BC (25. Cor.). Therefore, in the triangles In every right-angled triangle, the ABC, LMN, the three sides of the square of the hypotenuse, or side op one are equal to the three sides of the posite to the right angle, is equal to the other, each to each; and, consequently, sum of the squares of the sides which the angle B A C is (7.) equal to the angle contain that angle : and conversely, if MLN, that is, to a right angle.* the square of one side of a triangle be Therefore, &c. equal to the sum of the squares of the Cor. 1. In a right-angled triangle, the other two sides, the angle contained by square of either of the two sides is equal those two sides shall be a right angle. to the difference of the squares of the Let A B C be a right-angled trian hypotenuse and the other side. gle, having Cor. 2. It appears, from the demonthe stration, that if a perpendicular be right angle, B AC: drawn from the right angle to the hypothe square of BC shall be equal to tenuse, the square of either side is equal the sum to the rectangle under the hypotenuse of the and segment adjacent to that side. And squares of BA, AC. conversely, if this be the case, the angle Upon BC, BA, describe the squares at A must be a right angle. For, if the CD, A E; produce rectangle BC, BP, be equal to the DB to meet E F, or E F produced, in G; Of B P from each (31. and 36.) the rect square of B A; then, taking the square and, through A, draw H K parallel to angle BP, PC, will be equal to the BD. Then, because the sides E B, B G, of square of AP; and therefore, adding the square of PC to each, the rectangle the angle E BG, are perpendicular respectively to the sides AB, B C, of the BC, CP, will be equal to the square of angle A B C, these two angles. (18.) are B P, and B C, CP, together, that is (30. AC: therefore the two rectangles, BC, equal to one another; and the right angle B E G is equal to B AC; there- Cor.) the square of BC, will be equal to the fore the two triangles EBG, ABC, squares of B A, AC, and the angle having two angles of the one'equal tó BAC will be a right angle (36.), Cor. 3. And hence it follows that, if two angles of the other, each to each; BAC be a right angle, the square of and the interjacent sides, E B, A B, equal the perpendicular on the hypotenuse to one another, are equal in every respect, and B G is equal to B C or BD * It is evident that the sides of the squares upon (5.). And, because the parallelogram A B, AC, which are opposite to A B, AC, would AG, and the square A E, are upon the meet, if produced, in the point H. The latter part of the demonstration is, accordingly, equally applicable same base, and between the same pa to shew that if upon the two sides A B, A C, of any rallels (24.), A G is equal to A E. Again, triangle A B C (right-angled or otherwise) any two because the parallelogram AG, and the parallelograms be described whose sides opposite to À B, A C, meet in a point H, and if, upon the base rectangle KB, are upon equal bases, BC of the triangle, a parallelogram be likewise BG, BD, and between the same pa- described, having its sides adjoining to the base equal and parallel to A H, the parallelogram upon the base rallels (25.), A G is equal to K B. There shall be equal to the suin of the parallelograms upon fore, also, A E, that is, the square of AB, the two sides. M E B P DK |