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Let A be the given point, AB the given line, and IKL the given angle.
From the vertex K, as a cen
tre, with any radius, describe the
tance AB, equal to KI, describe the indefinite arc BO; then take a radius equal to the chord LI, with which, from the point B as a centre, describe an arc cutting the indefinite arc BO, in D; draw AD; and the angle DAB will be equal to the given angle K.
For, the two arcs BD, LI, have equal radii, and equal chords; hence they are equal (Prop. IV.); therefore the angles BAD, IKL, measured by them, are equal.
To divide a given arc, or a given angle, into two equal parts.
First. Let it be required to divide the arc AEB into two equal parts. From the points A and B, as centres, with the same radius, describe two arcs cutting each other in D; through the point D and the centre C, draw CD: it will bisect the arc AB in A the point E.
For, the two points C and D are each equally distant from the extremities A and
B of the chord AB'; hence the line CD bi
sects the chord at right angles (Book I. Prop. XVI. Cor.); hence, it bisects the arc AB in the point E (Prop. VI.).
Secondly. Let it be required to divide the angle ACB into two equal parts. We begin by describing, from the vertex C as a centre, the arc AEB; which is then bisected as above. It is plain that the line CD will divide the angle ACB into two equal parts.
Scholium. By the same construction, each of the halves AE, EB, may be divided into two equal parts; and thus, by successive subdivisions, a given angle, or a given arc may be divided into four equal parts, into eight, into sixteen, and so on.
Through a given point, to draw a parallel to a given straight line.
Let A be the given point, and BC the given line.
From the point A as a centre, with a radius greater than the shortest distance from A to BC, describe the indefinite arc EO; from the point E as
a centre, with the same radius, describe the arc AF; make ED AF, and draw AD: this will be the parallel required.
For, drawing AE, the alternate angles AEF, EAD, are evidently equal; therefore, the lines AD, EF, are parallel (Book I. Prop. XIX. Cor. 1.).
Two angles of a triangle being given, to find the third.
Draw the indefinite line DEF; at any point as E, make the angle DEC equal to one of the given angles, and the angle CEH equal to the other: the remaining angle HEF will be the third angle required; be- D
cause those three angles are
together equal to two right angles (Book I. Prop. I. and
Two sides of a triangle, and the angle which they contain, being given, to describe the triangle.
Let the lines B and C be equal to TB TC
the given sides, and A the given an
Having drawn the indefinite line
DE, at the point D, make the angle
EDF equal to the given angle A;
then take DG=B, DH=C, and draw GH; DGH will be the
triangle required (Book I. Prop. V.).
A side and two angles of a triangle being given, to describe the triangle.
The two angles will either be both adjacent to the given side, or the one adjacent, and the other opposite: in the latter case, find the third angle (Prob. VII.); and the two adjacent angles will thus be known: draw the straight line D
DE equal to the given side at the point D, make an angle EDF equal to one of the adjacent angles, and at E, an angle DEG equal to the other; the two lines DF, EG, will cut each other in H; and DEH will be the triangle required (Book I. Prop. VI.).
The three sides of a triangle being given, to describe the triangle.
Let A, B, and C, be the sides. Draw DE equal to the side A; from the point E as a centre, with a radius equal to the second side B, describe an arc; from D as a centre, with a radius equal to the third side C, describe another arc intersecting the former in F; draw DF, EF; and DEF will be the triangle required (Book I. Prop. X.).
Scholium. If one of the sides were greater than the sum of the other two, the arcs would not intersect each other: but the solution will always be possible, when the sum of two sides, any how taken, is greater than the third.
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
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.
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.
point D, make the angle EDF==
DEGF will be the parallelogram required.
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.
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
ference pass through three given points A, B, C ; and also for describing a circumference, in which, a given triangle ABC shall be inscribed.