Imágenes de páginas
PDF
EPUB

SECTION III.

THE POINT O THEOREM.

19. Theorem.-If points P, Q, and R be taken on the sides of a triangle the circles AQR, BRP, and CPQ pass through a common point 0.

For let the circles AQR and BRP meet in O. Then since (Euc. III. 22) QOR=T-A and ROP=π-В, we have QOP=2π- (π-A)—(π-B)=A+B=π- C; therefore the quadrilateral POQC is cyclic.

[graphic][subsumed][subsumed][subsumed][subsumed]

The angles BOC, COA, AOB, subtended by the sides of the given triangle at O, are respectively A+P, B+Q, C+R, when O is within the triangle ABC.

For, applying Euc. I. 32 to the triangles BOC and COA, it follows that LAOB-C+CAO+CBO.

[blocks in formation]

where R denotes an angle of the triangle PQR. Similarly for the angles BOC and COA.

If O falls outside the triangle ABC these angular relations become somewhat modified. Take for example 0 within the angle C.

Then from the cyclic quadrilaterals QRAO and RPBO we have (Euc. III. 20)

LORP OBP and [ORQ=OAQ;

[blocks in formation]

But

and

=

A-BOC=ABO-ACO ...

ABO=RPO since PRBO is cyclic,

ACO QPO since PQCO is cyclic.

=

[blocks in formation]

(1)

(B)

It may be shown in the same manner that if the points P, Q, R are such that two of the angles P, Q of the triangle formed by them are greater than A and B respectively BOC=P-A,

and

COA=Q-B,.............

AOB-C-R.

(x)

Hence if a triangle PQR of given species be inscribed in a given one ABC, the circles AQR, BRP, and CPQ

pass through either of two fixed points, one of which subtends at the sides of ABC, angles A+P, B+Q, C+R, and the other A-P, B-Q, R-C, or P-A, Q-B, C—R, according as two of the angles of the given triangle are greater or less than the corresponding angles of the inscribed triangle.

20. Let PQR be a triangle of given species inscribed in ABC. We have seen that the point O is fixed, and therefore the lines AO, BO divide the angles of ABC into known segments. But the segments of A are equal to the base angles of the triangle QOR; similarly of B to the base angles of ROP, and of C to the base angles of POQ.

Hence each of the triangles POQ, QOR, ROP are given in species. Therefore as the inscribed triangle PQR varies in position OQR, ORP, OPQ remain constant in species, and OP: 0Q OR are constant ratios.

Again, since the triangle OPQ is fixed in species and one vertex O a fixed point; if P describes a line BC it follows that the locus of Q is also a line (CA). And generally, when one vertex of a figure of given species is fixed and any other vertex P or point invariably connected with it describe a locus, the remaining points Q... describe loci, which may be derived from P by revolving it through a known angle POQ and increasing or diminishing OP in the ratio of OQ : OP.

The loci thus described are similar, the ratio OP : OQ is termed their Ratio of Similitude and the point O the Centre of Similitude.

Thus since O is a point invariably connected with a variable inscribed triangle PQR of given species, the

ortho-centre, circum-centre, ex-centres, median point, etc., and all other points invariably connected with the triangle, describe right lines which can at once be constructed by the above method.

Moreover, we know that if O is fixed and P describes a circle, and the variable line or Radius Vector OP be divided in Q, in a given ratio, the locus of Q is a circle. Now if Q be turned around 0 through any given angle the locus is the same circle displaced through the same angle. Therefore if one vertex of a triangle of given species is fixed, and another vertex describe a circle, the remaining vertex and all other points invariably connected with it likewise describe circles.

EXAMPLES.

1. Having given the diagonals and angles of a quadrilateral ABCD, construct it.

[graphic][subsumed][subsumed]

[On one diagonal AC describe segments of circles containing angles respectively equal to B and D. Let ABCD be the required quadrilateral. Produce CD to Y and BC to X. Join BY and AY.

Then since the chord BY of a given circle subtends a given angle C it is of known length. The triangle ADY is also given in species; hence the following construction :-On BY describe a segment of a circle containing an angle C. The triangle ADY, of given species,

has one vertex Y fixed, another A describing the circle AYC, therefore the remaining vertex D describes a circle. Take Bas centre and BD as radius, and cut this locus in the point D; therefore, etc.*] 2. Required to place a parallelogram of given sides with its vertices on four concurrent lines (M'Vicker).

[Let ABCD be the parallelogram situated on the pencil O. ABCD. Through C and D draw parallels CP and DP to BO and 40 respectively. Join OP. By Ex. 1 the diagonals and angles of the quadrilateral CDPO are given; therefore, etc.]

21. When the triangle PQR is given in every respect, the triangles OPQ, OQR, ORP are completely determined; for in addition to their species we are given the sides PQ, QR, and RP, hence the sides OP, OQ, OR are easily determined. We have therefore four solutions, real or imaginary, to the problem :

Having given two triangles ABC and PQR to place either with its vertices on the corresponding sides of the other; for having determined the point O, the position of which depends altogether on the species of the triangles, we get the position of the vertex P by taking O as centre and OP as radius and describing a circle cutting BC.

22. When the line OP is perpendicular to BC, OQ and OR are therefore perpendiculars to CA and AB respectively, and the circle with O as centre and OP as radius touches BC. In this case the two solutions coincide, and PQR is the minimum triangle of given species that can be inscribed in ABC.

23. It is manifest that a given triangle ABC may be escribed to another PQR. For having determined the point O, the triangles BOC, COA, and AOB are given in species, and are therefore completely determined, since

* For other solutions see "Mathematics from the Educational Times," Vol. XLIV., p. 29, by D. Biddle and Rev. T. C. Simmons.

« AnteriorContinuar »