25. AB is the diameter of a semi-circle, and P a variable point on its arc; find when a . PA + b. PB is maximum, if a, b are given whole numbers.

26. A and B are fixed points, one inside and one outside a fixed circle (centre C): if BC, AC are respectively m and n times the radius, show that the points in which the circle through A, B, C cuts the fixed circle, determine the positions of P, Q, on the fixed circle, for which—

1o, m . PA + n. PB is minimum; and,

2o, m. QA~n. QB is maximum

C, P being on opposite, and C, Q on the same side of AB.

27. Find P in the bisector of angle A, of triangle ABC, so that the difference of the angles PBC, PCB is maximum; and show that then the sum of these angles is half the angle BAC.

NOTE-If AC < AB, let 1 from C on bisector meet AB in D; draw thro. B, D to touch bisector in T: then thro. B, T, C will cut bisector in P.

28. If the number of sides of a polygon is fixed, and its corners lie on fixed lines, show that when its perimeter is minimum the fixed lines bisect its angles externally.

NOTE-Use Theorem (11) (a).

29. Of all isoperimetrical polygons, of a given number of sides, that of maximum area is the equilateral.

30. If all the sides of a polygon, excepting one, are given in length, show that its area is maximum when the remaining side is the diameter of a semi

circle, whose arc goes through all the corners.

Deduce from this that if all the sides of a polygon are given in length its area is maximum when its corners are concyclic.

31. (1) Of all polygons of a given number of sides, inscribed in a given circle, show that the regular one has its area and perimeter maximum.

(2) Of all polygons of a given number of sides, described about a given circle, show that the regular one has its area and perimeter minimum.

32. In Theorem (19) if P is on a fixed line, instead of a fixed circle, show that similar results are true.

33. By means of the last Exercise, find when the ratio of two sides of a triangle is maximum, being given that the third side (or base) is fixed in magnitude and position, and also that—

1o, the vertex is on a fixed line bisecting the base; or

2o, one base angle (supposed acute) is fixed; or

3o, the area is fixed.

NOTE-Results are—1o, vertical λ right; 2°, other base ▲ right; 3o, differ

ence of bases right.

SECTION ii-CONCURRENCY AND COLLINEARITY.

THEOREM (1)—If X, Y, Z are points in the sides BC, CA, AB of a triangle ABC, such that the perpendiculars to the sides at these points are concurrent, then

(BX2 — CX2) + (CY2 – AY2) + (AZ2 — BZ3)

and conversely.

= 0;

Let P be the pt. of concurrence.

Then joining P with A, B, C, we have

(BX2 - CX2) + (CY2 – AY2) + (AZ2 – BZ3),

=

BP2

= 0.

For the converse; let 1s at X, Y meet in P.

Suppose PZ' to AB.

Then, by preceding part, (CY2 – AY2) + (AZ′2 — BZ'2)

(BX2 — CX2) +

(BX2 — CX3) +

= 0;

(CY2 — AY2) + (AZ2 — BZ3), by hypothesis. ... Z and Z' are the same point.

THEOREM (2)-(Ceva's) When three lines AX, BY, CZ, drawn from the corners A, B, C of a triangle ABC, to meet its opposite sides in X, Y, Z, are concurrent, then

.., compounding the ratios, and recollecting that the ratio compounded of reciprocal ratios is unity, we get

(AZ: ZB) (BX: XC) (CY: YA) = 1.

For the converse; let AX, BY cut in P; and suppose that CP meets AB in Z': then, by preceding part,

THEOREM (3)—(Menelaus') If three points X, Y, Z, lying respectively on the three sides BC, CA, AB of a triangle ABC, are collinear, then— (AZ: ZB) (BX: XC) (CY: YA) = 1;

Def. Any line drawn across a system of lines is called a transversal of that system.

Hence the preceding Theorem may be expressed thus-If a transversal cuts the sides (or sides produced) opposite corners A, B, C, of triangle ABC, in X, Y, Z, respectively, then—

(AZ: ZB) (BX: XC) (CY: YA) = 1;

and conversely, if this relation holds between the segments into which the sides of a triangle ABC are cut (externally or internally) in X, Y, Z, then XYZ is a transversal of the triangle.

Note-If in the above result, the antecedents of the ratios are exchanged

N

A

B

X

among themselves, or the consequents among them-
selves, the result is still true; or we may take the
reciprocals of all the ratios. In order to recollect
the way in which the letters, in the three preceding
Theorems, are to be taken, place them round a circle,
as in the accompanying figure, where the corners of
the are placed as A, B, C; and the points in the
sides respectively opposite to them as X, Y,
beginning at any point, take the successive letters,
going one way round.

Z;

then,

THEOREM (4)-(Desargues') If two triangles are so placed that their corners connect concurrently, then their corresponding sides intersect collinearly; and conversely.

A

B

Then ▲ Paß is cut by transversal ZBA,
.. (aZ:ZB) (6B: BP) (PA : Aa) = 1.

And A Pẞy is cut by transversal XCB,
:. (BX:Xy) (C:CP) (PB:BB) = 1.
Also A Pay is cut by transversal YCA,
.. (Y: Ya) (aA: AP) (PC: Cy) = 1.

.., compounding these ratios, and recollecting that compounds of reciprocal ratios are unity, we get

For the converse; if X, Y, Z (the respective intersections of BC, By, of AC, ay, and of AB, aß) are collinear, let yC, BB meet in P.

Then As BBZ, CyY have joins of corresponding corners concurrent, viz., BC, By, ZY, meeting in X.

.., by first part, intersections of BB, YC, of ZB, YC, and of ZB, Yy are collinear;

i. e. P, A, a are in one line.

Aa, BB, Cy are concurrent.

Def. Two triangles, related as in the preceding Theorem, are said to be in perspective; and the point of concurrency, and line of collinearity, are termed respectively, the centre and axis of perspective.

Note-The appropriateness of the term perspective will be readily seen, by any one who has a slight knowledge of perspective drawing, from the following consideration-which also gives a proof of the Theorem.

Let ABC, aẞy be in different planes, not ||, so that P is the vertex of a pyramid, of which ABC, aẞy are triangular sections.

Then planes of ABC, aẞy will intersect in a line (L say).
But AB, aß, being in one plane, and not ||, must meet.

.. their pt. of meeting is in L.

Simrly. BC, By meet in L; and so also do CA, ya.

And this remains true if plane of ABC is turned about L as a hinge.

Let it be turned into coincidence with plane of a By: then Desargues' Theorem follows.

Def. Any number of collinear points, when taken in connection with each other, is termed a range.

Def. Any number of concurrent lines, when taken in connection with each other, is termed a pencil: the separate lines are termed rays; and the point of concurrency is termed the focus of the pencil.