Fig. 66.

Fig. 68.

Fig. 69.

129. Corollary III. Every angle BAC (fig. 66), inscribed in a segment greater than a semicircle, is an acute angle; for it has for its measure the half of the arc BOC less than a semicircumference.

And every angle BOC, inscribed in a segment less than a semicircle, is an obtuse angle; for it has for its measure the half of the arc BAC greater than a semicircumference.

130. Corollary IV. The opposite angles A and C (fig. 68) of an inscribed quadrilateral ABCD are together equal to two right angles; for the angle BAD has for its measure the half of the arc BCD, and the angle BCD has for its measure the half of the arc BAD; hence the two angles BAD, BCD, taken together, have for their measure the half of the circumference; therefore their sum is equal to two right angles.

THEOREM.

131. The angle BAC (fig. 69), formed by a tangent and a chord, has for its measure the half of the arc AMDC, comprehended between its sides.

Demonstration. At the point of contact A draw the diameter AD; the angle BAD is a right angle (110), and has for its measure the half of the semicircumference AMD; the angle DAC has for its measure the half of DC; therefore BAD + DAC, or BAC, has for its measure the half of AMD plus the half of DC, or the half of the whole arc AMDC.

It may be demonstrated, in like manner, that CAE has for its measure the half of the AC, comprehended between its sides.

Fig. 70.

Problems relating to the two first Sections.

PROBLEM.

132. To divide a given straight line AB (fig. 70) into two equal parts.

Solution. From the points A and B, as centres, and with a radius greater than the half of AB, describe two arcs cutting each other in D; the point D will be equally distant from the points A and B; find in like manner, either above or below the line AB a second point E equally distant from the points A and

B; through the two points D and E draw the line DE; this line will divide the line AB into two equal parts in the point C.

For, the two points D and E being each equally distant, from the extremities A and B, they must both be in the perpendicular which passes through the middle of AB. But through two given points only one straight line can be drawn; therefore the line DE will be this perpendicular, which divides the line AB into two equal parts in the point C.

PROBLEM.

133. From a given point A (fig. 71), in the line BC, to erect a Fig. 71. perpendicular to this line.

Solution. Take the points B and C, at equal distances from A; and from B and C, as centres, with a radius greater than BA, describe two arcs cutting each other in D; draw AD, which will be the perpendicular required.

For the point D, being equally distant from B and C, must be in a perpendicular to the middle of BC (55); therefore AD is this perpendicular.

134. Scholium. The same construction will serve to make a right angle BAD at a given point A in a given line BC.

PROBLEM.

135. From a given point A (fig. 72) without the straight line BD, Fig. 72. to let fall a perpendicular upon this line.

Solution. From A, as a centre, with a radius sufficiently great, describe an arc cutting the line BD in two points B and D ; then. find a point E, equally distant from the points B and D (132), and draw AE, which will be the perpendicular required.

For the two points A and E are each equally distant from the points B and D; therefore the line AE is perpendicular to the middle of BD.

PROBLEM.

136. At a given point A (fig. 73) in the line AB, to make an angle Fig. equal to a given angle K.

Solution. From the vertex K, as a centre, with any radius, describe an arc IL meeting the sides of the angle, and from the point A, as a centre, with the same radius, describe an indefinite

Fig. 74.

Fig. 75.

Fig. 76.

arc BO; from B, as a centrs, with a radius equal to the chord LI, describe an arc cutting the arc BO in D; draw AD, and the angle DAB will be equal to the given angle K.

For the arcs BD, LI, have equal radii and equal chords; they are therefore equal (102), and the angle BAD = IKL.

PROBLEM.

137. To bisect a given arc or angle.

Solution 1. If it is proposed to bisect the arc AB (fig. 74); from the points A and B, as centres, with the same radius, describe two arcs intersecting each other in D; through the point D and the centre C draw CD, which will divide the arc AB into two equal parts in the point E.

For, since the points C and D are each equally distant from the extremities A and B of the chord AB, the line CD is perpendicular to the middle of this chord; therefore it bisects the arc ᎯᏴ (105).

2. If it is proposed to bisect the angle ACB; from the vertex C, as a centre, describe the arc AB, and complete the construction, as above described. It is evident that the line CD will bisect the angle ACB.

138. Scholium. By the same construction, we may bisect. each of the halves AE, EB, and thus, by successive subdivisions, we may divide an angle or arc into four, eight, sixteen, &c., equal parts.

PROBLEM.

139. Through a given point A (fig. 15), to draw a straight line parallel to a given straight line BC.

Solution. From the point A, as a centre, with a radius sufficiently great, describe the indefinite arc EO; from the point E, as a centre, with the same radius, describe the arc AF; take

ED=AF,

and draw AD, which will be the parallel required.

For, AE being joined, the alternate angles AEF, EAD, are equal; therefore AD, EF, are parallel (67).

PROBLEM.

140. Two angles A and B (fig. 76) of a triangle being given, to find the third.

Solution. Draw the indefinite line DEF; at the point E make the angle DECA, and the angle CEH B; the remaining angle HEF will be the third angle required; for these three angles are together equal to two right angles.

PROBLEM.

141. Two sides of a triangle B and C (fig. 77) being given, and Fig. 77. the angle A contained by them, to construct the triangle.

Solution. Draw the indefinite line DE, and make at the point D the angle EDF equal to the given angle A; then take DG = B, DHC, and draw GH; DGH will be the triangle required (36).

PROBLEM.

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

Solution. The two given angles will be either both adjacent to the given side, or one adjacent and the other opposite. In this last case, find the third angle (140); we shall thus have the two adjacent angles. Then draw the straight line DE (fig. 78) Fig. 78. equal to the given side, at the point D make the angle EDF equal to one of the adjacent angles, and at the point E the angle DEG equal to the other; the two lines DF, EG, will cut each other in H, and DEH will be the triangle required (38).

PROBLEM.

143. The three sides A, B, C (fig. 79), of a triangle being given, Fig. 79, to construct the triangle.

Solution. 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 the point D, as a centre, with a radius equal to the third side C, describe another arc cutting the former in F; draw DF, EF, and DEF will be the triangle required (41).

144. Scholium. If one of the sides be greater than the sum of the other two, the arcs will not cut each other; but the solution will always be possible, when each side is less than the sum of other two.

PROBLEM.

145. Two sides A and B of a triangle being given with the angle C opposite to the side B, to construct the triangle.

Solution. The problem admits of two cases. 1. If the angle Fig. 80. C (fig. 80) is a right angle, or an obtuse angle, make the angle

Fig. 81.

Fig. 82.

Fig. 83.

EDF equal to the angle C; take DE = A, from the point E, as a centre, and with a radius equal to the given side B, describe an arc cutting the line DF in F; draw EF, and DEF will be the triangle required.

It is necessary, in this case, that the side B should be greater than A, for the angle C being a right or an obtuse angle, it is the greatest of the angles of the triangle, and the side opposite must consequently be the greatest of the sides.

2. If the angle C (fig. 81) is acute, and B greater than A, the construction is the same, and DEF is the triangle required.

But if, while C (fig. 82) is acute, the side B is 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 situated on the same side of D; there are therefore two triangles DEF, DEG, which equally answer the conditions of the problem.

146. Scholium. The problem would be in every case impossible, if the side B were less than the perpendicular let fall from E upon the line DF.

PROBLEM.

147. The adjacent sides A and B (fig. 83) of a parallelogram being given together with the included angle C, to construct the parallelogram.

=

Solution. Draw the line DEA; make the angle FDE=C, and take DF = B; describe two arcs, one from the point F, as a centre, with the radius FG DE, and the other from the point E, as a centre, with the radius EGDF; to the point G, where the two arcs cut each other, draw FG, EG; and DEGF will be the parallelogram required.

For, by construction, the opposite sides are equal, therefore the figure is a parallelogram (86), and it is formed with the given adjacent sides and included angle.

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