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requires but a very slight acquaintance with the former science, to perceive at once, that, when the lines contain each of them some common part M a certain number of times exactly, these theorems are but so many examples of its rules of addition, subtraction, and multiplication. The perspicuity alone, however, which is displayed in the above expressions, will enable the uninitiated reader to form some notion of the advantages resulting from a more intimate union of the two sciences.

is equal to the rectangle B K (ax. 1.). And, in like manner, it may be shewn that the square of BC is equal to the rectangle K C. Therefore, the squares of AB, A C, together, are equal to the rectangles B K, KC together, that is, to the square of B C.

Next, let the square of B C be equal to the sum of the squares of BA, AC: the angle B A C shall be a right angle. For, let L M N be another triangle having its sides, LM, LN, equal to the sides AB, 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 M N is

Triangles.

PROP. 36. (Euc. i. 47, and 48.) In every right-angled triangle, the square of the hypotenuse, or side opposite to the right angle, is equal to the sum of the squares of the sides which contain that angle: and conversely, if the square of one side of a triangle be equal to the sum of the squares of the other two sides, the angle contained by those two sides shall be a right angle. Let A B C be a right-angled triangle, having the right angle, B AC: the square of BC shall be equal to the sum of the squares of BA, AC. Upon B C, BA, describe the squares CD, AE; produce

E

T

D K

M

DB to meet E F, or E F produced, in G; and, through A, draw ĤI K parallel to

BD.

Then, because the sides E B, B G, of the angle E B G, are perpendicular respectively to the sides AB, B C, of the angle A B C, these two angles (18.) are equal to one another; and the right angle BE G is equal to BAC; therefore the two triangles EBG, ABC, having two angles of the one equal to two angles of the other, each to each, and the interjacent sides, E B, A B, equal to one another, are equal in every respect, and BG is equal to B C or BD (5.). And, because the parallelogram AG, and the square A E, are upon the same base, and between the same parallels (24.), A G is equal to A E. Again, because the parallelogram A G, and the rectangle KB, are upon equal bases, BG, BD, and between the same parallels (25.), A G is equal to K B. Therefore, also, A E, that is, the square of AB,

equal to the squares of L M, LN, that is, to the squares of A B, A C, or to the square of B C, and M N is equal to B C (25. Cor.). Therefore, in the triangles ABC, LMN, the three sides of the other, each to each; and, consequently, one are equal to the three sides of the the angie B A C is (7.) equal to the angle MLN, that is, to a right angle.*

Therefore, &c.

square of either of the two sides is equal Cor. 1. In a right-angled triangle, the to the difference of the squares of the hypotenuse and the other side.

Cor. 2. It appears, from the demonstration, that if a perpendicular be drawn from the right angle to the hypotenuse, the square of either side is equal to the rectangle under the hypotenuse and segment adjacent to that side. And conversely, if this be the case, the angle at A must be a right angle. For, if the rectangle B C, BP, be equal to the of BP from each (31. and 36.) the rectsquare of BA; then, taking the square angle BP, PC, will be equal to the square of AP; and therefore, adding the square of PC to each, the rectangle BC, CP, will be equal to the square of AC: therefore the two rectangles, B C, BP, and BC, CP, together, that is (30. Cor.) the square of B C, will be equal to the squares of BA, A C, and the angle BAC will be a right angle (36.).

BAC be a right angle, the square of Cor. 3. And hence it follows that, if the perpendicular on the hypotenuse

It is evident that the sides of the squares upon

AB, AC, which are opposite to A B, A C, would meet, if produced, in the point H. The latter part of the demonstration is, accordingly, equally applicable to shew that if upon the two sides A B, AC, of any triangle ABC (right-angled or otherwise) any two parallelograms be described whose sides opposite to A B, AC, meet in a point H, and if, upon the base BC of the triangle, a parallelogram be bikewise described, having its sides adjoining to the base equal and parallel to AH, the parallelogram upon the base shall be equal to the sum of the parallelograms upin the two sides.

will be equal to the rectangle under the segments of the hypotenuse; and that, conversely, if this be the case, the angle BAC will be a right angle. (See the proof of Cor. 2.)

Scholium.

Among the different proofs which have been invented of this celebrated theorem, there is one of no little elegance, which has the advantage of pointing out in what manner the squares of the two sides may be dissected, so as to form, by juxta-position of their parts, the square of the hypotenuse. It will be readily apprehended from the following outline.

BAC is

a triangle, having the angle at A a right angle: upon the hypotenuse B C is described the square BCDE, and through the points D, E, straight lines are

drawn parallel to B A, C A, to meet one another in K, and AC, A B produced, in the points F, G. Then it may be shewn that the four right-angled triangles, upon the sides of the square BD, are equal to one another; and, therefore together equal to twice the rectangle BA, AC: also, that the figure AFKG is a square, and equal to the square of BA+AC: and hence it easily follows that the square of BC is equal to the squares of B A, A C (32. and ax. 3.). To shew the dissection of the squares, Ee and Cc are drawn paral lel to A B, to meet B b, which is drawn parallel to A C, in the points c, e: then Cb is equal to the square of A B, and Eb to the square of A C; and the former is divided into two, and the latter into three parts, which may be placed (as indicated by the divisions of BD,) so as to fill up the space BCDE, which is the square of BC.

PROP. 37. (Euc. ii. 12. and 13.) In every triangle, the square of the side which is opposite to any given angle, is greater or less than the squares of the sides containing that angle, by twice the rectangle, contained by either of these sides, and that part of it, which is intercepted between the perpendicular let fall upon it from the opposite angle, and the given angle: greater, when the given angle is greater than a right angle, and less, when it is less.

А. В

Let A B C be any triangle, and C one of its angles; and from the angle A to the opposite side BC, or BC produced, let there be drawn the perpen dicular A D: the square of A B shall be greater or less than the squares of A C, C B, by twice the rectangle BC, CD; greater, if C be greater than a right angle, and less if it be less.

B DC

DB

When C is greater than a right angle, the opposite side, AB, and the perpendicular, A D, must lie different sides upon of A C; since, otherwise, in the triangle ACD, one of the angles would be a right angle, and another greater than a right angle, which is impossible (8.) For the like reason, when C is less than aright angle, the opposite side, A B, and the perpendicular, AD, must lie upon the same side of A C.

Therefore, according as C is greater or less than a right angle, the line BD will be the sum, or the difference of BC, CD and (32. and 33.) the square of B D will be greater or less than the squares of B C, CD by twice the rectAdd to each the angle BC, CD. square of AD: therefore, the squares of B D, AD will be greater or less than the squares of B C, CD, A D, by twice the rectangle B C, C D. But the square of AB is equal (36.) to the squares of BD, AD, and the square of AC is equal to the squares of CD, A D. Therefore, the square of AB will be greater or less than the squares of BC, AC by twice the rectangle B C, CD. Therefore, &c.

Cor. Any angle of a triangle is greater or less than a right angle, according as the square of the side opposite to it is greater or less than the squares of the sides by which it is contained.

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or to the base produced; the difference of the squares of the sides shall be equal to the difference of the squares of the segments of the base, or of the base produced.

B

DO B CD

Let ABC be a triangle, and from the vertex A to the base BC, or BC produced, let there be drawn the perpendicular AD: the difference of the squares of AB, A C, shall be equal to the difference of the squares of BD, CD. For the square of BD is as much greater than the square CD, as the squares of B D, AD together are greater than the squares of C D, AD together, that is (36.), as the square of AB is greater than the square of A C.

Therefore, &c.

We may remark that, if the triangle be isosceles, the segments of the base will be equal; and, that, in other cases, the greater segment of the base is always adjoining to the greater side. (12. Cor. 1. and 2.)

PROP. 39.

In an isosceles triangle, if a straight line be drawn from the vertex to any point in the base, or in the base produced, the square of this straight line shall be less or greater than the square of either of the two sides, by the rectangle under the segments of the base, or of the base produced.

BD E

DB D

Let A B C be an isosceles triangle, having the side AB equal to the side AC, and from the vertex A to any point D in the base, or in the base produced, let there be drawn the line AD: the square of AD shall be less or greater than the square of A C, by the rectangle B D, D C.

Bisect BC in E, (post. 3.) and join AE. Then, DC is equal to the sum of DE and EC; and because B E is equal to EC, BD is equal to the difference of DE and E C. Therefore the rectangle BD, DC is (34.) equal to the difference of the squares of DE, E C. But the line AE is perpendicular to BC, because the base BC of the isosceles triangle is bisected in E (6. Cor. 3.). Therefore, (38.) the difference of the squares of DE, EC, is equal to the difference of the squares of AD, AC. Therefore (ax. 1.) the rectangle BD, DC, is equal to the difference

of the squares of AD, AC; that is, the square of AD is less or greater than the square of AC, by the rectangle BD, D ̊C. Therefore, &c.

PROP. 40.

In every triangle the squares of the two sides are together double of the squares of half the base, and of the straight line, which is drawn from the vertex to the bisection of the base.

ÂÂ

Let ABC be any triangle, and from the vertex A to D, the middle point of the base, let there be drawn the line AD. The squares of A B, A C, shall be together double of the squares of B D, D A.

B DECBD CE

From the point A to B C, or B C produced, draw the perpendicular A E (12.). Then BE is equal to the sum of BD, DE; and, because B D is equal to DC, E C is equal to the difference of B D, DE. Therefore, (35.) the squares of B E, E C are together double of the squares of BD, DE. And the square of EA, taken twice, is double of the square of E A. Therefore the squares of BE, EA, CE, EA, are together double of the squares of BD, DE, EA; that is (36.), the squares of BA, A C_are_together double of the squares of BD, DA.

When ACB is a right angle, the perpendicular AE coincides with the side A C. Therefore B D is equal to DE, and the square of B E is double of the squares of BD, DE (29. Cor. 2.) ; also the square of BE (36. Cor. 1.) is equal to the difference of the squares of BA, A E, that is, of BA, AC; and hence the squares of B A, A E, that is, of B A, AC, are together double of the squares of B D, DA, as before.

Therefore, &c.

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together greater than the squares of AC, BD, by four times the square of EF.

Join BF, FD. Then, because the base AC of the triangle B A C is bisected in F, the squares of A B, BC are (40.) equal to twice the squares of B F, AF: and for the like reason the squares of CD, DA are equal to twice the squares of DF, AF: therefore, the squares of A B, B C, CD, D A are together equal to twice the squares of BF, DF, together with four times the square of A F. But, because the base BD of the triangle FBD is bisected in E, the squares of BF, DF, taken twice, are (40.) equal to four times the squares of BE, EF: therefore the squares of AB, BC, CD, DA are together equal to four times the squares of A F, BE, together with four times the square of EF; that is, to the squares of AC, BD, (39. Cor. 2.) together with four times the square of EF.

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SECTION 7.-Problems.

Upon a piece of paper for a plane, with a pen, a ruler, and a pair of compasses, it is evident, that, first, a straight line may be drawn from any one point to any other point; 2ndly, a terminated straight line may be produced to any length in a straight line; 3dly, from the greater of two straight lines, a part may be cut off equal to the less; and 4thly, a circle may be described from any centre, and with any distance from that centre.

Although the paper be not an exact plane, nor the pen such as may serve to draw an exact line, these defects admit of being removed to any required degree, and do not, in the least, affect the accuracy of our conclusions, with regard to exact lines and planes. An edge which is a right line, or nearly so, (because it is the common section of two planes, see Book IV.) may be obtained by doubling over a piece of paper upon itself, and a right angle (def. 10.) by doubling over this edge upon itself. These are both useful upon occasion, especially the right angle, which is so frequently required in geometrical constructions that a case of instruments is commonly provided with one. Among practical mechanics it is known under the name of the square. Parallel lines also occur so frequently, that it is convenient to have a ruler expressly for drawing them, called a parallel ruler. This may be made of two or three different forms, the best of which is that of a flat ruler running

upon two equal rollers.

Operations so simple require no further notice.

PROP. 42. Prob. 1. (Euc. 1. 1.) To describe an equilateral triangle upon a given finite straight line A B. From the points A, B,

as centres, with the common radius A B, describe two circles, intersecting one another in C, and A join CA, CB.

E

B

Then, because CA, C B are each of them equal to A B (def. 24.), they are (ax. 1.) equal to one another, and the triangle CA B is equilateral. Therefore, an equilateral triangle has been described upon the given finite straight line AB, which was required to be done.

PROP. 43. Prob. 2. (Euc. i. 10.) To bisect a given finite straight line A B.

Upon either side of A B (42.) describe an equilateral triangle: join the vertices or summits C,D, and let CD cut AB in E.

Then, because the triangles CAD, CBD have the three sides of the one equal to the three sides of the other, each to each, (7.) they are equal in every respect, and the angle ACD is equal to BCD. Therefore, the triangles ACE, BCE, having two sides of the one equal to two sides of the other, each to each, and the included angles equal to one another, are equal in every respect (4), and A E is equal to EB. Therefore, &c.

Cor. By the same construction, a straight line may be drawn, which shall bisect any given straight line at right angles.

N. B. It is evident, from 6. Cor. 4., that the same end will be obtained by joining the point of intersection C of any two equal circles described from the centres A, B, and cutting one another above A B, with the point of intersection D of any other two equal circles described from the same centres, and cutting one another either above or below AB: for C, D will be the vertices of isosceles triangles upon the same base. This observation may be of use when one of the triangles is already described, as in the first method of Prob. 4.

PROP. 44. Prob. 3. (Euc. i. 11.) To draw a straight line at right angles to a given straight line AB, from a given point C in the same.

First method. point A, and make CB equal to CA: upon A B describe (42.) the equilateral triangle DA B, and join D C.

In CA take any from the centre C,

с

B

Then, because the triangles DCA, DCB have the three sides of the one equal to the three sides of the other, each to each, (7.) the angle DCA is equal to DCB; and they are adjacent angles; therefore each of them is a right angle (def. 10.), and D C is at right angles to A B.

Second method. Take any point D which is not in AE, and from the centre D, with the radius DC describe the circle AC E, cutting the line AB a second time in the point A join AD, and produce it to meet the circumference in E, and join EC. Then because in the triangle ECA, the line D C drawn from the angle C to the middle of the opposite side equal to half that side, the angle C is (19. Cor. 4.) a right angle, and CE is at right angles to A B. If the points A, C, coincide, DC is (12. Cor. 3.) at right angles to

A B.

CB

Third method. Take any straight line M, and make C A equal to four times M from the centre C, with a radius equal to three times M, describe a circle from the centre A, with a radius equal to five times M, describe a second circle, cutting the former in D, and join DC, DA. Then, because the square of five times M is (29. Cor. 3.) equal to the square of four times M, together with the square of three times M, the square of AD is equal to the squares of A C, CD; and, therefore, (36.) the angle ACD is a right angle, or DC is at right angles to A D.

Therefore, &c.

M

C

The last two methods are particularly convenient, when the given point C is near the edge of the paper.

PROP. 45. Prob. 4. (Euc. i. 12.) To draw a straight line at right angles to a given straight line AB, from a given point C without it.

First method. Take any point D upon the other side of A B, and join CD:

with the radius CD, describe a circle cutting A B in the points A, B: bisect (43.) AB in E, and join C E.

E D

Then, because the triangles CEA, CEB have the three sides of the one equal to the three sides of the other, each to each, (7.) the angle CEA is equal to CEB, and (def. 10.) C E is at right angles to A B.

Second method. Draw from C to (43.) in D: from the centre D, with the AB any straight line CA: bisect CA radius DA or DC, describe a circle, cutting AB in a second point E, and join CE. Then, for the same reason, as in the second method of the preceding problem, CE is at right angles to AB. Also, if the points A, E coincide, CA is at right angles to A B.

D

E

Third method. In A B take any two points A, B, and from A, B, as centres, with the radii A C, BC, describe circles cutting one another a second time in F: join CF, and let CF cut AB in E. Then, because ACF, BCF, are isosceles triangles upon the same base CF, the line AB which

joins their summits, bisects the base at right angles (6. Cor. 4.); that is, CE is at right angles to A B.

Therefore, &c.

When CA is equal to CB, (which may happen, when the points A, B are on different sides of E,) the last of these three methods is, in practice, nearly the same as the first. The last two methods are applicable to the case, in which the point C falls near the edge of the paper.

PROP. 46. Prob. 5. (Euc. i. 9.) To bisect a given rectilineal angle BAC. In A B take any point B: make AC equal to A B: join BC: upon BC describe (42.) the equilateral triangle BD C, and join A D. Then, because the triangles A B D, ACD have the three sides of the one equal to the three sides of the other, each to each, (7.) the angle B A D is equal to the angle CAD.

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