To Bisect an Arc or Angle. To draw a Line Parallel to a given one.

the point A as a centre, by the preceding problein, make an arc BC equal to IL. Draw AC, and we have A K.

Demonstration. For the angles A and K being, by art. 100, measured by the equal arcs BC and IL, are equal.

137. Problem. To bisect a given arc AB (fig. 74).

Solution. Find a point D at equal distances from A and B. Through the point D and the centre C draw the line CD, which bisects the arc AB at E.

Demonstration. Draw the chord AB. Since the points D and C are at equal distances from A and B, the line DC is, by art. 132, perpendicular to the middle of the chord AB, and therefore by art. 116, it passes through the middle E of the arc AB.

138. Problem. To bisect a given angle A (fig. 75).

Solution. From A as a centre, with any radius, describe an arc BC, and, by the preceding problem, draw the line. AE to bisect the arc BC, and it also bisects the angle A.

Demonstration. The angles BAE and EAC are equal, for they are measured by the equal arcs BE and EC.

139. Problem. Through a given point A (fig. 76), to draw a straight line parallel to a given straight line BC.

Solution. Join EA, and, by the preceding problem, draw AD, making the angle EAD: AEF, and AD is parallel

to BC, by art. 31.

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140. Problem. Two angles of a triangle being given, to find the third.

To construct a Triangle and Right Triangle.

Solution. Draw the line ABC (fig. 77). At any point B draw the line BD, making the angle DBC equal to one of the given angles, and draw BE, making EBD equal to the other given angle, and ABE is the required angle.

Demonstration. For these three angles are, by art. 25, together equal to two right angles.

141. Problem. Two sides of a triangle and their included angle being given, to construct the triangle.

Solution. Make the angle A (fig. 78) equal to the given angle, take AB and AC equal to the given sides, join BC, and ABC is the triangle required.

142. Problem. One side and two angles of a triangle being given, to construct the triangle.

Solution. If both the angles adjacent to the given side are not given, the third angle can be found by art. 140.

Then draw AB (fig. 78) equal to the given side, and draw AC and BC, making the angles A and B equal to the angles adjacent to the given side, and ABC is the triangle required.

143. Problem The three sides of a triangle being given, to construct the triangle.

Solution. Draw AB (fig. 78) equal to one of the given sides, and, by art. 128, find the point C at the given distances AC and BC from the point C, join AC and BC, and ABC is the triangle required.

144. Scholium. The problem is impossible, when one of the given sides is greater than the sum of the other two.

145. Problem. To construct a right triangle, when a leg and the hypothenuse are given.

To construct a Parallelogram. To find the Centre of a Circle.

Solution. Draw AB (fig. 79) equal to the given leg. At A erect the perpendicular AC, from B as a centre, with a radius equal to the given hypothenuse, describe an arc cutting AC at C. Join BC, and ABC is the triangle required.

146. Problem. The adjacent sides of a parallelogram and their included angle being given, to construct the parallelogram.

Solution. Make the angle A (fig. 80) equal to the given angle, take AB and AC equal to the given sides, find the point D, by art. 128, at a distance from B equal to AC, and at a distance from C equal to AB. Join BD and DC, and ABCD is, by art. 79, the parallelogram required.

147. Corollary. If the given angle is a right angle, the figure is a rectangle; and, if the adjacent sides are also equal, the figure is a square.

148. To find the centre of a given circle or of a given arc.

Solution. Take at pleasure three points A, B, C (fig. 81) on the given circumference or arc; join the chords AB and BC, and bisect them by the perpendiculars DE and FG; the point O in which these perpendiculars meet is the centre required.

Demonstration. For, by art. 116, the perpendiculars DE and FG must both pass through the centre, and it must therefore be at their point of meeting.

Scholium. By the same construction a circle may be found, the circumference of which passes through

To draw a Tangent to a Circle.

three given points not in the same straight line, or in which a given triangle is inscribed.

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

Solution. a. If the given point A (fig. 82) is in the cir cumference, draw the radius CA, and through A draw AD perpendicular to CA, and AD is, by art. 120, the tangent required.

b. If the given point A (fig. 83) is without the circle, join it to the centre by the line AC; upon AC as a diameter describe the circumference AMCN, cutting the given circumference in M and N; join AM and AN, and they are the tangents required.

Demonstration. For the angles AMC and ANC are right angles, being inscribed in semicircles, and therefore AM and AN are perpendicular to the radii MC and NC at their extremities, and are, consequently, tangents, by art. 120.

150. Corollary. The two tangents AM and AN are equal; for the right triangles AMC and ANC are equal, by art. 63, since they have the hypothenuse AC common, and the leg MC equal to the leg NC, and, therefore, the other legs AM and AN are equal.

151. Problem. To inscribe a circle in a given triangle ABC (fig. 84).

Solution. Bisect the angles A and B by the lines AO and BO, and their point of intersection O is the centre of the required circle, and the perpendicular OD let fall from 0 upon the side AC is its radius.

Demonstration. The perpendiculars OD, OE, and OF

To inscribe a Circle in a Triangle.

let fall from O upon the sides of the triangle are equal to each other. For, in the right triangles OAD and OAE the hypothenuse OA is common; the angle OAD=OẠE by construction; and the third angle AOD = AOE, by art. 66; the triangles are, therefore, equal, by art. 53; and OD is equal to OE. In the same way it may be proved that OF OD = OE.

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Hence the circumference DFE passes through the points D, F, E, and the sides are tangents to it, by art. 120.

152. Corollary. The three lines AO, BO, and CO, which bisect the three angles of a triangle, meet at the same point.

153. Problem. Upon Upon a given straight line AB (figs. 85 and 86), to describe a segment capable of containing a given angle, that is, a segment such that each of the angles inscribed in it is equal to a given angle.

Solution. Draw BF, making the angle ABF equal to the given angle. Draw BO perpendicular to BF, and OC perpendicular to the middle of AB. From O, the point of intersection of OB and OC, with a radius OB ОА, describe the circumference BMAN, and BMA is the segment required.

Demonstration. Since BF is perpendicular to BO, it is a tangent to the circle, and therefore the angles AMB and ABF are equal, since they are each, by arts. 106 and 121, measured by half the arc ANB.

154. Scholium. If the given angle were a right angle, the segment sought would be a semicircle described upon the diameter AB.