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the same manner. Such division of a segment is called harmonic division.

.. the bisectors of any angle of a triangle divide the opposite side harmonically.

209°. Theorem.-A line through the vertex of a triangle dividing the base into parts which are proportional to the conterminous sides is a bisector of the vertical angle. (Converse of 208°.)

Let the line through B cut AC internally in F.

Then, BD
AF

being the internal bisector

AB
BC

AD

=

(208°), and

AB

=

by

DC

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But AD is < AF while DC is > FC.

.. the relation is impossible unless F and D coincide, i.e., the line is the bisector AD.

Similarly it may be proved that if the line divides the base externally it is the bisector AD'.

210°. Theorem.--The tangent at any point on a circle and the perpendicular from that point upon the diameter divide the diameter harmonically.

AB is divided harmonically in
M and T.

Proof.-LCPT=<PMT=7, (110°)
ДСРМУДСТР,

A

M

B

[blocks in formation]

CM CB

or

CB CT

AM AT

or

CB-CM CT-CB' MB BT'

.. AB is divided harmonically in M and T.

211°. The following examples give important results.

(195°, 2)

q.e.d.

Ex. 1. L, M, and N are tangents which touch the circle at

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Again, let L and N intersect in V. Then

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(176°, Cor. 2)

.......(A)

(114°, Cor. 1)

Similarly

BY=PR, .. PQ. PR=PC2. .........................(B)

........

Ex. 2. AD is a centre-line and DQ a perpendicular to it,

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Conversely, if Q moves so that the constant, the locus of Q is a line through A.

AP.AQ remains to the centre-line

Now, let the dotted lines represent rigid rods of wood or metal jointed together so as to admit of free rotation about the points A, C, P, U, V, and Q, and such that UPVQ is a rhombus (82°, Def. 1), and AU=AV, and AC=CP, AC being fixed.

PQ is the right bisector of UV, and A is equidistant from U and V. Therefore A, P, Q are always in line.

Also, PUQ is an isosceles triangle and UA is a line to the base, therefore UA2-UP2=AP.AQ (174°). But, UA and UP being constants, AP .AQ is constant.

And AC being fixed, and CP being equal to AC, P moves on the circle through A having C as centre.

.. Q describes a line to AC.

This combination is known as Peaucellier's cell, and is interesting as being the first successful attempt to describe a line by circular motions only.

Ex. 3. To construct an isosceles triangle of which each basal angle shall be double the vertical angle.

Let ABC be the triangle required, and let AD bisect the LA.

Then B=LBAD=2DAC, and LC is common to the triangles ABC and DAC. Therefore these E triangles are similar, and the ACAD is isosceles

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And BC is divided into extreme and mean ratio at D (183°,

Ex. 1). Thence the construction is readily obtained.

Cor. 1. The isosceles triangle ADB has each of its basal angles equal to one-third its vertical angle.

Hence

Cor. 2. LABC= 36°, <BAC=72°, LBDA=108°. (1) Ten triangles congruent with ABC, placed side by side with their vertices at B, form a regular decagon. (132°)

(2) The bisectors AD and CE of the basal angles of the AABC meet its circumcircle in two points which, with the three vertices of the triangle, form the vertices of a regular pentagon.

(3) The LBDA=the internal angle of a regular pentagon.

211. The following Mathematical Instruments are imP portant:

B

1. Proportional Compasses.

This is an instrument primarily for the purpose of increasing or diminishing given line-segments in a given ratio; i.e., of multiplying given linesegments by a given tensor.

If AO BO and QO-PO, the triangles AOB, POQ are isosceles and similar, and

AB: PQ OA: OP.

Hence, if the lines are one or both capable of rotation about O, the distance AB may be made to vary at pleasure, and PQ will remain in a constant ratio to AB.

The instrument usually consists of two brass bars with slots, exactly alike, and having the point of motion O so arranged as to be capable of being set at any part of the slot. The points A, B, P, and Q are of steel.

P

5

2

B

2. The Sector.

This is another instrument which primarily serves the purpose of increasing or diminishing given line-segments in given ratios.

This instrument consists of two rules equal in length and jointed at O so as to be opened and shut like a pair of compasses. Upon each rule various lines are

drawn corresponding in pairs, one on each rule.

Consider the pair OA and OB, called the "line of lines." Each of the lines of this pair is divided into 10 equal parts

which are again subdivided. Let the divisions be numbered from 0 to 10 along OA and OB, and suppose that the points numbered 6 are the points P and Q. Then OAB and OPQ are similar triangles, and therefore PQ: AB=OP: OA. But OP=AO. .. PQ=AB. And as by opening the instrument AB may be made equal to any segment not beyond the compass of the instrument, we can find PQ equal to of any such given segment.

The least consideration will show that AB, 3-3 is AB, etc. Also that 3-3 is 7-7, etc.

the distance 5–5 is

of 7–7, 5–5 is § of Hence the instrument serves to divide any given

segment into any number of equal parts, provided the number is such as belongs to the instrument.

The various other lines of the sector serve other but very similar purposes.

3. The Pantagraph or Eidograph. Like the two preceding instruments the pantagraph pri- m, marily increases or diminishes segments in a given ratio, but unlike the others it is so ar- K ranged as to be continuous in its operations, requiring only one setting and no auxiliary instruments.

It is made of a variety of

forms, but the one represented

E

O

in the figure is one of the most convenient.

N

B

AE, AB, and BF are three bars jointed at A and B. The bars AE and BF are attached to the wheels A and B respectively, which are exactly of the same diameter, and around which goes a very thin and flexible steel band C.

The result is that if AE and BF are so adjusted as to be parallel, they remain parallel however they be situated with respect to AB. E, F are two points adjustable on the bars

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