PROBLEM X. Given, the three sides of a triangle, to construct the triangle. Let A, B, and C, be the given sides. Draw DE, and make it equal D AH B E to the side A; from D as a centre, with a radius equal to the side B, describe an arc; from E as a centre, with a radius equal to the side C, describe an arc intersecting the former at F; draw DF and EF: then will DEF be the triangle required (B. I., P. X.). Scholium. In order that the construction may be possible, any one of the given sides must be less than the sum of the other two, and greater than their difference (B. I., P. VII., S.). PROBLEM ΧΙ. Given, two sides of a triangle, and the angle opposite one of them, to construct the triangle. Let A and B be the given sides, and C the given angle. Draw an indefinite line DG, and at some point of it, as D, construct an angle GDE equal to the given angle; on one side of this angle lay off the distance DE equal to the side B adjacent to the given angle; from E as BH E D a centre, with a radius equal to the side opposite the given angle, describe an arc cutting the side DG at G; draw EG. Then will DEG be the required triangle. For, the sides DE and EG are equal to the given sides, and the angle D, opposite one of them, is equal to the given angle. opposite the given angle is side, there will be but one A Scholium. When the side greater than the other given solution. When the given angle is acute, and the side apposite the given angle is less than the other given side, and greater than the shortest distance from E to DG, there will be two solutions, DEG and DEF. When the side opposite the given angle is BH D equal to the shortest distance from E to DG, the arc will be tangent to DG, the angle opposite DE will be a right angle, and there will be but one solution. When the side opposite the given angle is shorter than the distance from E to DG, there will be no solution. PROBLEM XII. Given, two adjacent sides of a parallelogram and their included angle, to construct the parallelogram. Let A and B be the given sides, and C the given angle. allel to DF · then will DFGE be the parallelogram re. quired. For, the opposite sides are parallel by construction; and consequently, the figure is a parallelogram (D. 28); it is also formed with the given sides and given angle. PROBLEM XIII. To find the centre of a given circumference. Take any three points A, B, and C, on the circumference or arc, and join them by the chords AB, BC; bisect these chords by the perpendiculars DE and FG: then will their point of intersection O, be the centre required (P. VII.). Scholium. The same construc * C B tion enables us to pass a circumference through any three points not in a straight line. If the points are vertices of a triangle, the circle will be circumscribed about it. PROBLEM XIV. Through a given point, to draw a tangent to a given circle. There may be two cases: the given point may lie on the circumference of the given circle, or it may lie without the given circle. 1o. Let C be the centre of the given circle, and A a point on the circumference, through which the tangent is to be drawn. Draw the radius CA, and at A draw AD perpendicular to AC: then will AD be the tangent required (P. IX.). D 2o. Let C be the centre of the given circle, and 4 a point without the circle, through which the tangent is to be drawn. Draw the line AC; bisect it at 0, and from 0 as a centre, with a radius OC, describe the circumference ABCD; join the point A with the points of intersection D and B: then will both AD and AB be tangent to the given circle, and there will be two solutions. For, the angles ABC and ADC are right angles (P. XVIII., C. 2): B hence, each of the lines AB and AD is perpendicular to a radius at its extremity; and consequently, they are tangent to the given circle (P. IX.). Corollary. The right-angled triangles ABC and ADC, have a common hypothenuse AC, and the side BC equal to DC; and consequently, they are equal in all their parts (B. I., P. XVII.): hence, AB is equal to AD, and the angle CAB is equal to the angle CAD. gents are therefore equal, and the line AC bisects the angle between them. The tan let fall the perpendiculars OD, OE, OF, on the sides of the triangle these perpendiculars will all be equal. For, in the triangles BOD and BOE, the angles OBE and OBD are equal, by construction; the angies ODB and OEB are equal, because both are right angles; and consequently, the angles BOD and BOE are also equal all (B. I., P. XXV., C. 2), and the side OB is common; and their parts (B. I., In like manner, it From 0 as a centre, with a radius OD, describe a circle, and it will be the circle required. For, each side is perpendicular to a radius at its extremity, and is therefore tangent to the circle. Corollary. The lines that bisect the three angles of a triangle all meet in one point. PROBLEM XVI. On a given straight line, to construct a segment that shall contain a given angle. Produce AB towards D; at B construct the angle DBE equal to the given angle draw BO perpendicular |