PQ. But the only point common to these lines is 0, their point of intersection.

.. O is the required point.

5. Proof. By joining OA, OB, OC, these three lines may be shewn equal to one another by I. 4.

The point is evidently the centre of the circle passing through A, B, and C.

The following may stand as a further example :

To determine a point whose distances from two intersecting lines shall have given magnitudes.

1. Let AB, CD be the given lines intersecting in O. And let E be the distance of the point from AB and F its distance from CD.

is the locus in which the required point must lic. (ii.) 3. Similarly if OH be drawn perpendicular to CD and F the point will lie in the line HQ drawn parallel to CD.

ii.

4. But Q is the only point common to these lines. .. Qis the point required.

5. Proof. By dropping perpendiculars from Q on AB and CD, the proof is completed by prop. 34, I.

If the perpendiculars are drawn on the opposite side of AB and CD, the point q will also fulfil the conditions.

Hence, there are two points which satisfy the conditions of the problem.

6. GENERAL REMARKS.-In the solution of problems it will frequently be found, as in the example just given, that more than one solution is afforded by a given construction. In other cases a solution already discovered will be found to fail when the data have certain conceivable relations to one another. These cases are highly instructive, and should be carefully examined so as to determine how many solutions the problem admits of, and under what circumstances it becomes impossible. Thus, we may easily find a point in a given line, which shall be equally distant from two given points. But if the line joining the points is perpendicular to the given line, no point in this line will be equally distant from the two given points; and on examination we find that the construction fails, and we also see why it fails. A further example will be found on page 44 of the School Euclid, and this, with the one above given, should be carefully studied. But the consideration of these exceptional cases should always be reserved until the solution of the possible cases has been obtained.

The methods above given will only be of use to diligent and persevering pupils. No method will be of any service to the student who has not fully mastered the propositions on the solution of which the exercises depend, and who will not use independent effort and be resolved to think for himself. Even such an one must be prepared for many failures and disappointments, and after adopting first one method and then another the solution will often occur to him as a happy thought which comes he knows not how. But every effort, even if apparently unsuccessful, prepares the mind to seize the right method when it actually emerges even if it does not actually suggest it. Finally, a little ingenuity and common sense will often aid us more than the best methods, and we will therefore conclude with an example which will illustrate this.

A ball is placed on a rectangular billiard table. In what direction must it be struck so that after being reflected from the four sides in succession it may return to the point from which it set out?

N.B. It may be assumed that the angles which the ball makes with a side on striking and on leaving it are equal to one another.

In the figure, ABCD represents the table O the initial position of the ball, and POQRSP its path.

Now producing RQ backwards to meet the perpendicular OX in E, the 4° XQE= RQC=OQX, and the 4 at X are right / and XQ common; hence, XE=0X, [I. 26], and OE=2 0X.

Similarly, if SR be produced to meet the perpendicular to DC from E in F, then KF=EK=XC.

In like manner LG=LF=DK, and lastly HM= MG=AL.

Now H being thus determined, HO may be joined and produced to Q; then if EQ be joined and produced to R, FR may then be joined and produced to S; and, finally, GS when produced will give SP.

Thus the path is determined. The synthetical proof may be left to the pupil.

EXERCISES.-BOOK I.

Sides and Angles of Triangles. Conditions of Equality.

Props. 4, 8, 26, 5 and 6.

1. The perpendiculars from the middle points of two sides of a triangle meet in a point which is equidistant from the three angles of the triangle.

2. Find a point within an acute-angled triangle from which a circle may be described about the triangle.

3. Two angles of an equilateral triangle are bisected, and the point at which the bisecting lines intersect is joined with the third angle; show that the third angle will be bisected.

4. The exterior angles at the base of an isosceles triangle ABC are bisected by the straight lines AO, BO intersecting in O. If CO be joined show that it will bisect the vertical angle ACB.

5. In any isosceles triangle if a straight line be drawn from the vertex to the base and fulfil any one of the following conditions it must fulfil the other two. The conditions are, that the line (1) bisects the base, (2) bisects the vertical angle, (3) is perpendicular to the base. (Three exercises.)

[These properties of an isosceles triangle form the key to the next three exercises.]

6. To find a point in a given straight line which shall be equidistant from two given points.

7. Through a given point to draw a straight line equally inclined to two given lines.

8. From two given points on opposite sides of a line draw two straight lines which shall meet in that given straight line and include an angle bisected by the given straight line.

9. The diagonals of a rhombus bisect each other at right angles.

10. Inscribe a rhombus in a given triangle, having one of its angles coincident with an angle of the triangle.

11. Show that the perpendiculars dropped from any point in the line bisecting an angle of a triangle upon the sides containing that angle, or those sides produced, will be equal to one another.

12. Find a point in a given straight line from which the perpendiculars upon two given straight lines which intersect shall be equal to one another.

Sides and Angles of Triangles. Conditions of Inequality.

Props. 18, 19, 20, 21, 24, 25.

13. If AD is the longest side and BC the shortest of a quadrilateral ABCD, then the angle ABC is greater than ADC, and the angle BCD greater than BAD.

14. In any quadrilateral figure show that any one side is less than the sum of the other three.

15. The four sides of a quadrilateral are together greater than the sum and less than twice the sum of the diagonals.

16. The two sides of a triangle are together greater than twice the line joining the vertex with the middle point of the base.

17. The sum of the straight lines drawn from any point within a triangle to each of its angles is less than the sum and greater than half the sum of the sides of the triangle.

18. Show that the last exercise is true if a square be substituted for the triangle.

19. Two quadrilateral figures ABCD and AEFD stand on the same base AD and AEFD lies wholly within ABCD. Show that the perimeter of AEFD is less than that of ABCD.

[Produce EF to meet AB and CD in G and H.]