APOLLONIUS' THEOREM.

Ex. 47. Use Apollonius' theorem to calculate the length of the median of an equilateral triangle of side 2 in. Verify by Pythagoras' theorem.

Ex. 48. Use Apollonius' theorem to calculate the lengths of the three medians in a triangle whose sides are 4, 6, 7.

Ex. 49. Repeat Ex. 48, (i) with sides 4, 5, 7, (ii) with sides a, b, c.

Ex. 50. Calculate the base of a triangle whose sides are 8 cm. and 16 cm., and whose median is 12 cm. Verify graphically.

Ex. 51. The base BC of an isosceles ▲ ABC is produced to D, so that CD=BC; prove that AD2=AC2+2BC2.

Ex. 52. A side PR of an isosceles ▲ PQR is produced to S so that RS-PR; prove that QS2=2QR2+ PR2.

TEX, 53. What does Apollonius' theorem become if the vertex moves down (i) on to the base, (ii) on to the base produced?

Ex. 54. A point moves so that the sum of the squares of its distances from two fixed points A, B remains constant; prove that its locus is a circle having for centre the mid-point of AB.

Ex. 55. The base AD of a triangle OAD is trisected in B, C. Prove that OA2+20D2=30C2+6CD2.

ន

(Apply Apollonius' theorem to ▲ OAC, OBD; then eliminate OB2.)

Ex. 66. In the figure of Ex. 55, OA2+OD2=OB2+OC2+4BC2.

Ex. 57. The sum of the squares on the sides of a parallelogram is equal to the sum of the squares on the diagonals.

Ex. 58. In any quadrilateral the sum of the squares on the four sides exceeds the sum of the squares on the diagonals by four times the square on the straight line joining the mid-points of the diagonals.

(Let E, F be the mid-points of AC, BD; apply Apollonius' theorem to ▲ BAD, BCD and AFC.)

Ex. 59. The sum of the squares on the diagonals of a quadrilateral is equal to twice the sum of the squares on the lines joining the mid-points of opposite sides. (See Ex. 44 on p. 51 and Ex. 57 above.)

Ex. 60. In a triangle, three times the sum of the squares on the sides =four times the sum of the squares on the medians.

GEOMETRICAL RIDERS REDUCING TO ALGEBRAICAL IDENTITIES.

The following exercises can be solved by representing lengths by symbols and using Algebraic methods.

Ex. 61. M is the mid-point of AB and O is any point on AB or AB produced. Prove that OM (OA+OB) or † (OA~OB), distinguishing the two cases.

=

Ex. 62. PQ is a straight line bisected at V; PQ is produced to O. Show that OV2 - PV2=OP.OQ.

Ex. 63. A straight line is divided into two equal and also into two unequal parts; prove that the difference of the squares on the two unequal parts is equal to twice the rectangle contained by the whole line and the part between the points of section.

Ex. 64. ACDB is a straight line, and D bisects CB; prove that the square on AC is less than the sum of the squares on AD, DB by twice the rectangle AD.DB.

Ex. 65. If a straight line be bisected and produced to any point, the square on the whole line thus produced is equal to the square on the part produced together with twice the rectangle contained by the line and the line made up of the half and the part produced.

Ex. 66. AB is divided into any two parts in C, and AC, BC are bisected in D, E; prove that the sum of the square on AE and three times the square on BE is equal to the sum of the square on BD and three times the square on AD.

Ex. 67. A, B, C, D are four points in order in a straight line. Prove that AB. CD+AD. BC=AC.BD.

CHAPTER IX

LOCI

To show that the locus of a point moving under specified conditions is a certain line or lines (straight or curved) which we will call 7, we must prove two things: (i) that every point satisfying the conditions lies on 7, (ii) that every point on 7 satisfies the conditions.

The need of care in the matter is illustrated by the following problem. AB, AC are equal straight lines; it is required to find the locus of points at which AB, AC subtend equal angles.

RB

Fig. 71.

It is evident from symmetry that every point on the unlimited straight line PQ bisecting BAC satisfies the conditions. But it cannot be proved that every point satisfying the conditions lies on PQ. In fact, the locus consists of PQ together with circle ABC and the lines BR and Cs. These parts of the locus might have escaped notice if we had been content to prove (ii) without examining (i) also.

the arc BDC of the

THEOREM 34. The locus of a point which is equidistant from two fixed points is the perpendicular bisector of the straight line joining the two fixed points. (i) Data A and B are fixed points. P is any

one position of a point such that PA = PB.

To prove that Plies on the perpendicular bisector

of AB.

Construction Join AB; let N be the middle point of AB. Join NP.

N

UNV is the perpendicular bisector of AB and P any point on it.

(ii) Data

..any point on the perpendicular bisector is equidistant from A and B.

Q. E. D.

NOTE. It will be noticed that N is a point on the locus.

THEOREM 35. The locus of a point which is equidistanț from two intersecting straight lines consists of the pair of straight lines which bisect the angles between the two given lines.

B

M

*

Fig. 73.

B

(i) Data AOA', BOB' are two intersecting straight lines.

=

P is any point such that PM PN where PM and PN are perpendiculars from P to AOA', BOB'.

To prove that P lies on one of the bisectors of the angles formed by

AOA', BOB'.

.. P lies on the bisector of one of the angles formed by AOA'

Sim' if PO bisects any of the angles formed by AOA', BOB' it may be shown that P is equidistant from AOA' and BOB'.

Q. E. D.

Ex. 1. Write out the proof of Th. 35 for a figure drawn with P in the LAOB'. In each of Exx. 2-6 there are two things to be proved, see p. 72.

Ex. 2. If O is a fixed point and P moves along a straight line, prove that the locus of the mid-point of OP is the right bisector of the perpendicular from O on to the given line.

Ex. 3. A variable line moves so that its mid-point remains fixed and one extremity of the line moves along a fixed line. Prove that the locus of the other extremity is a parallel to the given line.