41. From a point there can

only be dropped one perpendic

ular to a line, and this is shorter than any other line from the point to the line.

When BC AC, then we cannot have BAAC, as there cannot be two right angles in a triangle (18); further by 38, BA > BC as CL A.

B

A

C.

The point C, where the perpendicular from B meets the line AC, is called the projection of B on AC. If D is the projection of another point E, CD is called the projection of the line BE.

42. Two lines which diverge equally

from the perpendicular are equal.

That AB and BC diverge equally either signifies, that x= Zy or that

B

xy

43. When two lines diverge unequally from the perpendicular, that is greatest which diverges

most.

If both lines lie at the same side of the perpendicular, the proposition follows from 38, as x> <y; if they lie one on each side, the one is turned round to the same side as the other.

44. Every point in the perpendicular on the middle point of a line is equidistant from the extremities of the line.

4

B

B

୯

AB

The proposition follows from 42, as

BC.

45. A point, lying outside the perpendicular on the middle point of a line, is nearest to that extremity of the line, which is on the same side.

For if the perpendicular BD be drawn, then AD > DC, therefore AB > BC (43). The perpendicular on the middle point of a line therefore just contains all the points equidistant from the extremities of the line, and no others.

46. If two triangles have two sides equal, but the angles contained by them unequal, the third side is greatest in the triangle which has the greatest angle.

D

E

B

Let ACB and ADB be the triangles; they are placed, so that the equal sides AB coincide, further AD =AC; a perpendicular on the middle of CD will pass through A (45, last piece); therefore CB > DB. (45). 47. The altitude of a triangle is a line, drawn from an angular point, perpendicular to the opposite side, which is then called the base. The altitude must coincide with one of the sides, if one of the angles at the base is a right angle, and fall outside the triangle, if one of the angles is obtuse; besides this line, two others of particular interest issue from each angular point, namely, the median line, going to the middle of the opposite side, and the line bisecting the angle.

In the isosceles triangle the altitude, median line and the line bisecting the vertical angle, coincide in one line; for if the altitude did not bisect the vertical angle or the base, the sides of the triangle would be unequal (43).

48. The middle point of a chord, the middle points of the corresponding arcs, and the centre all lie in a line, perpen-· dicular on the middle point of the chord. For all the points lie equidistant from the extremities of the chord.

EXAMPLES.

48. Prove that the diameter of a circle is the greatest chord. 49. Prove that the arcs between two parallel chords in a circle are equal.

50. Which is the greatest and which the least line, that can be drawn from a point to the circumference of a circle? 51. From a point 0 within a triangle ABC lines are drawn to B and C. Prove that OB + OC < AB+ AC. 52. Prove that in a circle, when AB <~ AC, then also ~ AB AC. (The arcs are both supposed less than 180°). (46).

53. Prove that each side of a triangle is less than half the perimeter.

54. Prove that in a circumscribed quadrilateral, the sum of the one pair of opposite sides is equal to the sum of the other pair. (29).

55. With a point A outside a circle as centre, a circle is described through the centre 0 of the given circle, and two chords are drawn from 0, equal to the diameter of the given circle. Shew that these chords cut the given circle in the points of contact of the tangents from A.

56. Shew that any triangle, which lies wholly inside another triangle, has less perimeter than this.

57. From a point within a triangle lines are drawn to the

three angular points; shew that the sum of these three lines is greater than half the perimeter of the triangle, but less than the whole perimeter.

58. From the angular points of a triangle three lines are drawn in such a manner, that any two of them are legs of an isosceles triangle, of which one of the sides is the base. Prove that the three lines intersect in the same point. (Suppose that the three lines form a triangle and express the sides of this in terms of the legs of the isosceles triangles).

II.

1. CONSTRUCTION, CONGRUENCE AND SYMMETRY.

49. For the construction of figures with certain given properties are used only the ruler, by which a straight line can be drawn through two given points, and the compasses, by which a circle, with given centre and given radius, can be described. All constructions must therefore be a combination of these two. In order that a point may be determined, there must be given two conditions, which it must fulfil; if there is only one condition given, there will be an infinite number of points, which satisfy the problem, but these will all lie in a certain straight or curved line, called the locus of the point; thus we have from the preceding:

The locus of the points which are at a given distance from a given point is a circle, with the given point as centre and the given distance as radius.

The locus of the points which are equidistant from two given points is a straight line, perpendicular on the middle point of the line joining the two given points. (45).

More loci will be mentioned hereafter.

Now if the problem is to find a point, we examine, which two loci correspond to the two conditions, which the point, according to its position, must fulfil; if the two loci can be constructed by compasses and ruler, the point can be found,

for, as it is to lie in them both, it must lie where they intersect; the problem has therefore as many solutions, as the loci have points of intersection.

50. When certain parts only in one way can be put together to make a figure, then two figures, both containing these parts, must be congruent, for if they were not, the parts would be put together in two ways; but if the parts can be put together in different ways, then two figures containing these parts do not require to be congruent.

51. Two figures are said to be symmetrical with regard to a straight line, when to each point in the one figure, there is a corresponding point in the other, so placed, that it is found by dropping a perpendicular from the first point on to the line, and producing it equally far on the other side; thus A corresponds to a, B to h, &c. Symmetrical figures are congruent, as the one, by being turned over, round the straight line (axis of symmetry), will coincide with the other; conversely, two figures, which, by being turned over, round a straight line, would coincide, must lie symmetrically with regard to it. As examples of symmetrical figures, we may mention that the altitude from the vertex divides an isosceles triangle symmetrically, that every diameter divides a circle symmetrically, &c.

A

a

D

C

B

b

d

When two pair of figures lie symmetrically with regard to the same line, their points of intersection must also lie symmetrically, for when the figures, by being turned, coincide, their points of intersection must also coincide.

In constructing a figure, we often get two symmetrical solutions, which therefore are congruent.

52. To draw a straight line, perpendicular on the middle point of a given straight line.

With the extremities A and B of the line given as centres, describe arcs with equal radii, then their points of intersection must lie in the line required (45); in this manner two points in the required line are found, which is then drawn through these.