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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 Eas 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.
Secondly. If the angle C is acute, and B greater than A, the same construction will again apply, and DEF will be the triangle required.
But if the angle C is acute, and the side B less than A, then the arc described from the centre E, with the radius EF=B, will cut the side DF in two points F and G, lying on the same side of D: hence there will be two triangles DEF, DEG, either of which will satisfy the conditions of the problem.
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. 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 centre, with a radius FG-DE, the D other from E as a centre, with a radius EG=DF; to the point A G, where these arcs intersect B each other, draw FG, EG; 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.
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.).
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.
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.
Through a given point, to draw a tangent to a given circle.
If the given point A lies in the circumference, 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.
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.
Let ABC be the given triangle. Bisect the angles A and B, by the lines AO and BO, meeting in the point 0; from the point O, let fall the perpendiculars OD, OE, OF, on the three sides of the triangle: these perpendiculars will all be equal. For, by construc
To inscribe a circle in a given triangle.
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.
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.