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tion, we have the angle DAO OAF, the right angle ADO= AFO; hence the third angle AOD is equal to the third AOF (Book I. Prop. XXV. Cor. 2.). Moreover, the side AO is common to the two triangles AOD, AOF; and the angles adjacent to the equal side are equal: hence the triangles themselves are equal (Book I. Prop. VI.); and DO is equal to OF. In the same manner it may be shown that the two triangles BOD, BOE, are equal; therefore OD is equal to OE; therefore the three perpendiculars OD, OE, OF, are all equal.

Now, if from the point O as a centre, with the radius OD, a circle be described, this circle will evidently be inscribed in the triangle ABC; for the side AB, being perpendicular to the radius at its extremity, is a tangent; and the same thing is true of the sides BC, AC.

Scholium. The three lines which bisect the angles of a triangle meet in the same point.

PROBLEM XVI.

On a given straight line to describe a segment that shall contain a given angle; that is to say, a segment such, that all the angles inscribed in it, shall be equal to the given angle.

Let AB be the given straight line, and C the given angle.

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Produce AB towards D; at the point B, make the angle DBE C; draw BO perpendicular to BE, and GO perpendicular to AB, through the middle point G; and from the point O, where these perpendiculars meet, as a centre, with a distance OB, describe a circle: the required segment will be AMB.

For, since BF is a perpendicular at the extremity of the radius OB, it is a tangent, and the angle ABF is measured by half the arc AKB (Prop. XXI.). Also, the angle AMB, being an inscribed angle, is measured by half the arc AKB: hence we have AMB ABF=EBD=Č: hence all the angles inscribed in the segment AMB are equal to the given angle C.

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Scholium. If the given angle were a right angle, the required segment would be a semicircle, described on AB as a diameter.

PROBLEM XVII.

To find the numerical ratio of two given straight lines, these lines being supposed to have a common measure.

Let AB and CD be the given lines.

From the greater AB cut off a part equal to the less CD, as many times as possible; for example, twice, with the remainder BE.

From the line CD, cut off a part equal to the remainder BE, as many times as possible; once, for ample, with the remainder DF.

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From the first remainder BE, cut off a part equal to the second DF, as many times as possible; once, for example, with the remainder BG.

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From the second remainder DF, cut off a part equal G to BG the third, as many times as possible.

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Continue this process, till a remainder occurs, which is contained exactly a certain number of times in the preceding one.

Then this last remainder will be the common measure of the proposed lines; and regarding it as unity, we shall easily find the values of the preceding remainders; and at last, those of the two proposed lines, and hence their ratio in numbers.

Suppose, for instance, we find GB to be contained exactly twice in FD; BG will be the common measure of the two proposed lines. Put BG=1; we shall have FD=2: but EB contains FD once, plus GB; therefore we have EB=3: CD contains EB once, plus FD; therefore we have CD=5: and, lastly, AB contains CD twice, plus EB; therefore we have AB=13; hence the ratio of the lines is that of 13 to 5. If the line CD were taken for unity, the line AB would be; if AB were taken for unity, CD would be.

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Scholium. The method just explained is the same as that employed in arithmetic to find the common divisor of two numbers: it has no need, therefore, of any other demonstration.

How far soever the operation be continued, it is possible that no remainder may ever be found, which shall be contained an exact number of times in the preceding one. When this happens, the two lines have no common measure, and are said to be incommensurable. An instance of this will be seen after

wards, in the ratio of the diagonal to the side of the square. In those cases, therefore, the exact ratio in numbers cannot be found; but, by neglecting the last remainder, an approximate ratio will be obtained, more or less correct, according as the operation has been continued a greater or less number of times.

PROBLEM XVIII.

Two angles being given, to find their common measure, if they have one, and by means of it, their ratio in numbers.

Let A and B be the given angles.

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B

E

F

With equal radii describe the arcs CD, EF, to serve as measures for the angles: proceed afterwards in the comparison of the arcs CD, EF, as in the last problem, since an arc may be cut off from an arc of the same radius, as a straight line from a straight line. We shall thus arrive at the common measure of the arcs CD, EF, if they have one, and thereby at their ratio in numbers. This ratio will be the same as that of the given angles (Prop. XVII.); and if DO is the common measure of the arcs, DAO will be that of the angles.

Scholium. According to this method, the absolute value of an angle may be found by comparing the arc which measures it to the whole circumference. If the arc CD, for example, is to the circumference, as 3 is to 25, the angle A will be of four right angles, or 1 of one right angle.

It may also happen, that the arcs compared have no common measure; in which case, the numerical ratios of the angles will only be found approximatively with more or less correctness, according as the operation has been continued a greater or less number of times.

BOOK IV.

OF THE PROPORTIONS OF FIGURES, AND THE MEASUREMENT

OF AREAS.

Definitions.

1. Similar figures are those which have the angles of the one equal to the angles of the other, each to each, and the sides about the equal angles proportional.

2. Any two sides, or any two angles, which have like positions in two similar figures, are called homologous sides or angles.

3. In two different circles, similar arcs, sectors, or segments, are those which correspond to equal angles at the centre. Thus, if the angles A and O are equal, the arc BC will be similar to DE, the sector BAC to the sector DOE, and the segment whose chord is BC, to the segment whose chord is DE.

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4. The base of any rectilineal figure, is the side on which the figure is supposed to stand.

5. The altitude of a triangle is the perpendicular let fall from the vertex of an angle on the opposite side, taken as a base. Thus, AD is the altitude of the triangle BAC

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6. The altitude of a parallelogram is the perpendicular which measures the distance between two opposite sides taken as bases. Thus, EF is the altitude of the parallelo- A gram DB.

7. The altitude of a trapezoid is the perpendicular drawn between its two parallel sides. Thus, EF is the altitude of the trapezoid DB.

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F B

DE C

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8. The area and surface of a figure, are terms very nearly synonymous. The area designates more particularly the superficial content of the figure. The area is expressed numeri

cally by the number of times which the figure contains some other area, that is assumed for its measuring unit.

9. Figures have equal areas, when they contain the same measuring unit an equal number of times.

10. Figures which have equal areas are called equivalent. The term equal, when applied to figures, designates those which are equal in every respect, and which being applied to each other will coincide in all their parts (Ax. 13.): the term equivalent implies an equality in one respect only; namely, an equality between the measures of figures.

PROPOSITION I. THEOREM.

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We may here premise, that several of the demonstrations are grounded on some of the simpler operations of algebra, which are themselves dependent on admitted axioms. Thus, if we have A=B+C, and if each member is multiplied by the same quantity M, we may infer that Ax M=BxM+CxM; in like manner, if we have, A=B+C, and D=E-C, and if the equal quantities are added together, then expunging the +C and C, which destroy each other, we infer that A+D=B+ E, and so of others. All this is evident enough of itself; but in cases of difficulty, it will be useful to consult some agebraical treatise, and thus to combine the study of the two sciences.

Parallelograms which have equal bases and equal altitudes, are equivalent.

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Let AB be the common base of D CF EDF CE the two parallelograms ABCD, ABEF: and since they are supposed to have the same altitude, their upper bases DC, FE, will be both situated in one straight line parallel to AB.

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Now, from the nature of parallelograms, we have AD=BC, and AF=BE; for the same reason, we have DC=AB, and FE AB; hence DC-FE: hence, if DC and FE be taken away from the same line DE, the remainders CE and DF will be equal: hence it follows that the triangles DAF, CBE, are mutually eqilateral, and consequently equal (Book I. Prop. X.).

But if from the quadrilateral ABED, we take away the triangle ADF, there will remain the parallelogram ABEF; and if from the same quadrilateral ABED, we take away the equal triangle CBE, there will remain the parallelogram ABCD

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