Fig. 60.

Now produce AC and take CE = EF=FG.. .. = AC or BD. Draw DE, DE, DG....cutting AB in P3, P4, P5,,......

Then from similar triangles,

P3B:AP3 = BD:AE.

... P3B: AB = BD: AF = 1:3. Similarly

P4B:AB = 1 :4, and so on.

If AB = 1,

AP2 = 1 / 1.2;

P2P3 = 1 / 2.3;

P3P4 = 1 / 3.4;

........................

PnPn+1 = 1 / n(n+1).

But AP2+ P2P3 +P3P4+...is ultimately = AB.

... 1 /1.2 + 1 / 2.3 + 1 / 3.4 + ......to ∞ = 1. Or

1 - 1/2 = 1/ 1.2;

1/2 - 1/3 = 1/2.3;

.................................

1/n - 1 / n +1 = 1 /n(n+1).

Adding

1 / 1.2 + 1 / 2.3 + ...... + 1 / n(n+1) = 1 - 1 / n+1.

... 1 / 1.2 + 1 / 2.3 ...... + 1 / (n-1)n = 1 - 1 / n.

The limit of 1 - 1/n when n is ∞ is 1.

173. The following simple contrivance may be used for dividing a line into a number of equal parts.

Take a rectangular piece of paper, and mark off n equal segments on each or one of two adjacent sides. Fold through the points of section so as to obtain perpendiculars to the sides. Mark the points of section and the corners 0, 1, 2,.. .. n. Suppose it is required to divide the edge of another piece of paper AB into n equal parts. Now place AB so that A or B may lie on 0, and B or A on the perpendicular through n.

In this case AB must be greater than ON. But the smaller side of the rectangle may be used for smaller lines.

The points where AB crosses the perpendiculars are the required points of section.

174. Center of mean position. If a line AB contains (m + n) equal parts, and it is divided at C so that AC contains m of these parts and CB contains n of them; then if from the points A, C, B perpendiculars AD, CF, BE be let fall on any line, m . BE + n . AD = (m + n) . CF. Now, draw BGH parallel to ED cutting CF in G and AD in H. Suppose through the points of division AB lines are drawn parallel to BH. These lines will divide AH into (m + n) equal parts and CG into n equal parts.

... n-AH=(m + n).CG, and since DH and BE are each = GF, n.HD + m.BE=(m + n) GF. Hence, by addition n.1 + D + m.BE = (m+n) GF n.AD + m. BE = (m+ n) . CF C is called the center of mean position, or the mean center of A and B for the system of multiples m and n.

The principle can be extended to any number of points, not in a line. Then if P represent the feet of the perpendiculars on any line from A, B, C, etc., if a, b, c.....be the corresponding multiples, and if M be the mean center a.AP+b.BP+c.CP.. ..

= (a+b + c+.. ..).MP. If the multiples are all equal to a, we get a(AP+BP+CP+...)=na.MP n being the number of points.

175. The center of mean position of a number of points with equal multiples is obtained thus. Bisect the line joining any two points A, B in G, join G to a third point C and divide GC in H so that GH=1/3GC; join H to a fourth point D and divide HD in K so that HK = ¼HD and so on: the last point found will be the center of mean position of the system of points.

176. The notion of mean center or center of mean position is derived from Statics, because a system of material points having their weights denoted by a, b, c.. .., and placed at A, B, C.. . .would balance about the mean center M, if free to rotate about M under the action of gravity.

The mean center has therefore a close relation to the center of gravity of Statics.

177. The mean center of three points not in a line, is the point of intersection of the medians of the triangle formed by joining the three points. This is also the center of gravity or mass center of a thin triangular plate of uniform density.

178. If M is the mean center of the points A, B, C, etc., for the corresponding multiples a, b, c, etc., and if P is any other point, then a.AP2+b.BP2 + c.CP2 + ....

= a.AM2 + b.BM2 + c.CM2 + .. ..

Hence in any regular polygon, if O is the in-center or circum-center and P is any point

AP2 + BP2 +....= OA2 + OB2 +..... + n. OP2

= n.(R2+ OP2).

Now

AB2 + AC2 +AD2 + .. .. = 2n.R2.

Similarly

BA2 + BC2 + BD2 + .. .. = 2n.R2 CA2+ CB2+ CD2+.....=2n.R2.

Adding

2(AB2 + AC2 + AD2 +....) = n.2n.R2.

... AB2 + AC2 + AD2 + .. .. = n2.R2.

179. The sum of the squares of the lines joining the mean center with the points of the system is a minimum.

If M be the mean center and P any other point not belonging to the system,

∑PA2 = ∑MA2+∑PM2, (where ∑ stands for "the sum of all expressions of the type").

... ∑PA2 is the minimum when PM=0, i. e., when P is the mean center.

180. Properties relating to concurrence of lines and collinearity of points can be tested by paper folding.* Some instances are given below:

(1) The medians of a triangle are concurrent. The common point is called the centroid.

(2) The altitudes of a triangle are concurrent The common point is called the orthocenter.

(3) The perpendicular bisectors of the sides of a triangle are concurrent. The common point is called the circum-center.

(4) The bisectors of the angles of a triangle are concurrent. The common point is called the in-center.

(5) Let ABCD be a parallelogram and P any point. Through P draw GH and EF parallel to BC and AB respectively. Then the diagonals EG, HF, and the line DB are concurrent.

*For treatment of certain of these properties see Beman and Smith's New Plane and Solid Geometry, pp. 84, 182.

(6) If two similar unequal rectineal figures are so placed that their corresponding sides are parallel, then the joins of corresponding corners are concurrent. The common point is called the center of similarity.

(7) If two triangles are so placed that their corners are two and two on concurrent lines, then their corresponding sides intersect collinearly. This is known as Desargues's theorem. The two triangles are said to be in perspective. The point of concurrence and line of collinearity are respectively called the center and axis of perspective.

(8) The middle points of the diagonals of a complete quadrilateral are collinear.

(9) If from any point on the circumference of the circum-circle of a triangle, perpendiculars are dropped on its sides, produced when necessary, the feet of these perpendiculars are collinear. This line is called Simson's line.

Simson's line bisects the join of the orthocenter and the point from which the perpendiculars are drawn.

(10) In any triangle the orthocenter, circum-center, and centroid are collinear.

The mid-point of the join of the orthocenter and circum-center is the center of the nine-points circle, so called because it passes through the feet of the altitudes and medians of the triangle and the mid-point of that part of each altitude which lies between the orthocenter and vertex.

The center of the nine-points circle is twice as far from the orthocenter as from the centroid. This is known as Poncelet's theorem.

(11) If A, B, C, D, E, F, are any six points on a circle which are joined successively in any order, then the intersections of the first and fourth, of the second and fifth, and of the third and sixth of these joins produced when necessary) are collinear. This is known as Pascal's theorem.

(12) The joins of the vertices of a triangle with the points of contact of the in-circle are concurrent. The same property holds for the ex circles.

(13) The internal bisectors of two angles of a triangle, and the external bisector of the third angle intersect the opposite sides collinearly.

(14) The external bisectors of the angles of a triangle intersect the opposite sides collinearly.

(15) If any point be joined to the vertices of a triangle, the lines drawn through the point perpendicular to those joins intersect the opposite sides of the triangle collinearly.

(16) If on an axis of symmetry of the congruent triangles ABC, A'B'C a point O be taken A'O, B'O, and C'O intersect the sides BC, CA and AB collinearly.

(17) The points of intersection of pairs of tangents to a circle at the extremities of chords which pass through a given point are collinear. This line is called the polar of the given point with respect to the circle.

(18) The isogonal conjugates of three concurrent lines AX, BX, CX with respect to the three angles of a triangle ABC are concurrent. (Two lines AX, AY are said to be isogonal conjugates with respect to an angle BAC, when they make equal angles with its bisector.)

(19) If in a triangle ABC, the lines AA', BB', CC' drawn from each of the angles to the opposite sides are concurrent, their isotomic conjugates with respect to the corresponding sides are also concurrent. (The lines AA', AA" are said to be isotomic conjugates, with respect to the side BC of the triangle ABC, when the intercepts BA' and CA" are equal.)

(20) The three symmedians of a triangle are concurrent. (The isogonal conjugate of a median AM of a triangle is called a symmedian.)