such a triangle is impossible. (7.1.) But the solution will always be possible, when the sum of two sides is greater than the third side. PROBLEM XI. Two sides A and B of a triangle, and the angle C opposite the side B, being given, to describe the triangle. There are two cases. First. When the angle C is a right or an obtuse angle, make the angle In this first case, the side B must D be greater than A; for the angle C, C F being a right, or an obtuse angle, is the greatest angle of the triangle, therefore the side opposite to it, must also be the greatest of the sides. (13. 1.) Second. If the angle C is acute, and B greater than A, the same construction will again apply, and DEF will be the triangle required. AH E BH will be two triangles DEF, DEG, either of which will satisfy the conditions of the problem. Scholium. 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 A and B of a parallelogram, with the angle C which they contain, being given, to describe the parallelogram. Draw the line DE=A; at the point D, make the angle FDE=C; take DFB; describe two arcs, the one from F as a centre, with a radius FG-DE, the other from E as a centre, with a radius EG DF; to the point G, where D A B these arcs intersect 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: (31. 1 :) and it is formed with the given sides and the given angle. Cor. If the given angle be 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. G* E To find the centre of a given circle or arc. Take three points, A, B, C, anywhere in the circumference, or the arc; join AB, BC, or suppose them to be joined; bisect those two lines by the perpendiculars DE, FG: the point O, where these perpendiculars meet, will be the centre sought. (6.2. Sch.) F Scholium. The same construction serves for making a circumference 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. right-angle; (13. 2. Cor. 2;) therefore AB is a perpendicular at the extremity of the radius CB; hence it is a tangent. (9.2.) 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 rightangled triangles CBA, CDA have the hypothenuse CA common, and the side CB-CD; hence they are equal; (17.1;) therefore AD is equal to AB, and also the angle CAD to CAB. PROBLEM XV. To inscribe a circle in a given triangle ABC. A B F E C 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 construction, we have the angle DAO=OAF, the right-angle ADO=AFO; hence the third angle AOD is equal to the third AOF. (28.1. 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; (6. 1;) and DO is equal to OF. In the same manner it may be shown that the two triangles BOD, BOE are equal; hence 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 AB, to describe a segment containing a given angle C; that is to say, a segment such that all the angles inscribed in it shall be equal to the given angle C. Produce AB towards D; at the point B, make the acute, or obtuse angle DBE C; draw BO perpendicular to BE, and GO perpendicular to AB, bisecting it in G; and from the point O as a centre, where those perpendiculars meet, 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. (14. 2.) Also the angle AMB, being an inscribed angle, is measured by half the arc AKB: hence we have AMB=ABF=EBD=C: therefore all the angles inscribed in the segment AMB are equal to the given angle C. Scholium. If the given angle were a right-angle, the required segment would be the semicircle described on the diameter AB. |