PROBLEM XI.

Two sides of a triangle, and the angle opposite one of them, being given, to describe the triangle.

Let A and B be the given sides, and C the given angle. There are two cases.

First. When the angle C is a right angle, or when it is obtuse, make the angle EDF=C; take DE=A; from the point E as a centre, with a radius equal to the given side B, describe an arc cutting DF E in F; draw EF: then DEF will be the triangle required.

In this first case, the side B must be greater than A; for the angle C, being a right angle, or an obtuse an- D gle, is the greatest angle of the tri

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B

angle, and the side opposite to it must, therefore, also be the greatest (Book I. Prop. XIII.).

Scholium. If the arc described with E as a centre, should be tangent to the line DG, the triangle would be right angled, and there would be but one solution. The problem would be impossible in all cases, if the side B were less than the perpendicular let fall from E on the line DF.

PROBLEM XII.

The adjacent sides of a parallelogram, with the angle which they contain, being given, to describe the parallelogram

Let A and B be the given sides, and C the given angle.

Draw the line DE=A; at the

point D, make the angle EDF==
C; take DF-B; describe two
arcs, the one from F as a cen-
tre, with a radius FG-DE, the D
other from E as a centre, with
a radius EG=DF; to the point Ar
G, where these arcs intersect BH
each other, draw FG, EG;

T

DEGF will be the parallelogram required.

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For, the opposite sides are equal, by construction; hence the figure is a parallelogram (Book I. Prop. XXIX.): and it is formed with the given sides and the given angle.

Cor. If the given angle is a right angle, the figure will be a rectangle; if, in addition to this, the sides are equal, it will be a square.

PROBLEM XIII.

To find the centre of a given circle or arc.

Take three points, A, B, C, any where in the circumference, or the arc; draw AB, BC, or suppose them to be drawn; bisect those two lines by the perpendiculars DE, FG: the point O, where these perpendiculars meet, will be the centre sought (Prop. VI. Sch.).

Scholium. The same construction serves for making a circum

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ference pass through three given points A, B, C; and also for describing a circumference, in which, a given triangle ABC shall be inscribed.

PROBLEM XIV.

Through a given point, to draw a tangent to a given circle.

If the given point A lies in the circum- D ference, draw the radius CA, and erect AD perpendicular to it: AD will be the tangent required (Prop. IX.).

If the point A lies without the circle, join A and the centre, by the straight line CA: bisect CA in O; from O as a centre, with the radius OC, describe a circumference intersecting the given circumference in B; draw AB: this will be the tangent required.

For, drawing CB, the angle CBA being inscribed in a semicircle is a right angle (Prop. XVIII. Cor. 2.); therefore AB is a perpendicular at the extremity of the radius CB; therefore it is a tangent.

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A

Scholium. When the point A lies without the circle, there will evidently be always two equal tangents AB, AD, passing through the point A: they are equal, because the right angled triangles CBA, CDA, have the hypothenuse CA common, and the side CB=CD; hence they are equal (Book I. Prop. XVII.); hence AD is equal to AB, and also the angle CAD to CAB. And as there can be but one line bisecting the angle BAC, it follows, that the line which bisects the angle formed by two tangents, must pass through the centre of the circle.

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.

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 example, with the remainder DF.

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 o to BG the third, as many times as possible.

B

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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 3; if AB were taken for unity, CD would be.

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