126. Theorem.

Tangent Circumferences.'

When the circumferences of two

circles cut each other, the line AB (fig. 67), which. joins their centres, is perpendicular to the middle of the line CD, which joins their points of intersection.

Proof. For if a perpendicular be erected upon the middle of the chord CD, it must, by § 117, pass through the centres A and B of both the circles of which CD is a chord.

127. Theorem. When two circumferences are tangents to each other, their centres and point of contact are in the same straight line perpendicular to their common tangent at the point of contact.

Proof. a. If the centres of two circumferences which cut each other (fig. 67) are removed from each other, until the points C and D of intersection approach infinitely near to each other, the circles will become tangent, as in (fig. 56), the chord CD of (fig. 67) will become the tangent CD of (fig. 56); and as both the radii AM and MB are perpendicular to their common tangent, these radii must be in the same straight line.

b. In the same way, the centres of the circles (fig. 67) may be brought near to each other until the circles are tangents, as in (fig. 57), and the same reasoning may be here applied to prove that the line ABM, perpendicular to the common tangent at M, passes through both the centres A and B.

Position of a Point in a plane.

CHAPTER IX.

PROBLEMS RELATING TO THE FIRST EIGHT CHAPTERS.

128. Problem. To find the position of a point in a plane, having given its distances from two known points in that plane.

Solution. Let the known points be A and B (fig. 68). From the point A as a centre, with a radius equal to the distance of the required point from A, describe an arc. Also, from the point B as a centre, with a radius equal to the distance of the required point from B, describe an arc cutting the former arc; and the point of intersection C is the required point.

Scholium.

By the same process, another point D may also be found which is at the given distances from A and B, and either of these points therefore satisfies the conditions of the problem.

129. Corollary. If both the radii were taken of equal magnitudes, the points C and D thus found would be at equal distances from A and B.

130. Scholium. The problem is impossible, when the distance between the known points is greater than the sum of the given distances or less than their difference.

131. Scholium. If the required point is to be at equal distances from the known point, its distance from either of them must be greater than half the distance between the known points.

To Bisect a Line; to Erect a Perpendicular.

132. Problem. To divide a given straight line AB (fig. 69) into two equal parts; that is, to bisect it.

Solution. Find by § 129, a point C at equal distances from the extremities A and B of the given line. Find also another point D, either above or below the line, at equal distances from A and B. Through C and D draw the line CD, which bisects AB at the point E.

Proof. For the perpendicular, erected at E to the line AB, must, by § 42, pass through the points C and D, and must therefore, by § 16, coincide with the line CD.

133. Problem. At a given point A (fig. 70), in the line BC, to erect a perpendicular to this line.

Solution. Take the points B and C at equal distances from A; and find a point D equally distant from B and C. Join AD, and it is the perpendicular required.

Proof. For the point D must, by § 42, be a point of the perpendicular erected at A.

134. Problem. From a given point A (fig. 71), without a straight line BC, 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 BC in two points B and C ; find a point D equally distant from B and C, and the line ADE is the perpendicular required.

Proof. For the points A and D, being equally distant from B and C, must, by § 42, be in this perpendicular.

135. Problem. To make an arc equal to a given arc AB (fig. 72), the centre of which is at the given point C.

Solution. Draw the chord AB. From any point D as a centre, with a radius equal to the given radius CA,

To make and to bisect a given Arc, or Angle.

describe the indefinite arc FH. From F as a centre, with a radius equal to the chord AB, describe an arc cutting the arc FH in H, and we have the arc FH= AB.

=

Proof. For as the chord AB=the chord FH, it follows, from 112, that the arc AB = the arc FH.

=

136. Problem. At a given point A (fig. 73), in the line AB, to make an angle 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, by the preceding problem, make an arc BC equal to IL. Draw AC, and we have A K.

Proof. For the angles A and K being, by $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.

Proof. Draw the chord AB. Since the points D and C are at equal distances from A and B, the line DC is, by 132, perpendicular to the middle of the chord AB, and therefore by § 117, 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.

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

To construct a Triangle.

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 § 31.

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

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

Proof. For these three angles are, by § 25, together equal 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 § 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