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III. The third step in solving a Rider is to decide on the Method of Proof. The student's success here will depend on his mental ability, but there are several things which will materially help him to secure success.

In the first place, the results arrived at in Euclid must be thoroughly known. The enunciations of the Propositions in Euclid should be learnt by heart. A summary of the chief results arrived at in Props. 1—26 is given on page 52, and these results should be carefully learnt. An examination also of the Methods of Proof used by Euclid will help the student. This examination shows that Euclid's Proofs of Props. 1—26 may be arranged into five classes, but it is to be observed that:

First, Axiom I. is used to prove no less than five Propositions. The student may expect therefore to use this Axiom whenever he has to prove equality of two magnitudes.

Secondly, most of the propositions in Euclid depend for their proof on some other proposition, and Props. 4 and 8 are the propositions most used. In all such cases equality of some part of two triangles has to be proved. Furthermore :

(1) To prove equality of a side, Prop. 4 is generally used.
(2) To prove equality of an angle, Prop. 8 is generally used, and

(3) To prove equality of two sides, Prop. 26 can be used ; or we may remember that

(1) Prop. 4 is used when two sides and the included angle are given equal. (2) Prop. 8 is used when three sides are given equal, and (3) Prop. 26 is used when two angles and one side are given equal.

Also to prove some property in a single triangle, any of the Propositions 16 to 21 may be used, and Prop. 16 is the one most used.

Again, Props. 13, 14, 15 may be used to prove properties of angles.

But in many cases the drawing of the figure, and stating what is taken for granted and what has to be proved, will at once suggest some Proposition already proved in Euclid, as a model for proving the Rider.

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IV. If the student has failed to discover the right method of proof for any Rider by the above means, there are two other ways by which he may perhaps be led to find the proof.

He may begin by assuming or taking for granted what he is required to prove, and then trying what deductions he can make with this assumption. Frequently he will come to some result which will suggest the construction, or to some Theorem which he knows to be true. If then he reverses the process, he will probably have discovered the correct solution of the Rider. For example, suppose we are asked to prove in the figure of Prop. 16 that AB and BC are together greater than twice BE.

Begin by assuming that AB and BC are together greater than twice BE.

We know that AB is equal to CF, and that twice BE is equal to BF, therefore CF and BC are greater than BF; and we know that this is true by Prop. 20.

The proof therefore will run as follows:-
Because by Prop. 20 in the triangle BCF the two sides CF and BC are

together greater than the third side BF,
and because CF is equal to AB; and BF is equal to twice BE;

therefore AB and BC are together greater than twice BE. Q. E. D.

V. Or again, the solution of a Rider may often be suggested by drawing a number of different figures to illustrate the different cases. For example, suppose we are asked to find the locus of all points which are equidistant from two given points.

Let A and B be the two given points ; it is required to find the locus of all points equidistant from A and B. Mark on the figure a number of points, each one of A

B which is equidistant from A and B. The Succession of points suggests a straight line for the required locus ; and as the middle point of the line joining A and B is on the locus, we easily sce that the required locus is a straight line bisecting the straight line AB at right angles.

This last device is mostly useful in problems. It will generally be found that it is possible to draw certain extreme or critical cases, and these are often the most suggestive. For example, in the Rider just given, the point which bisects the line joining AB is a critical case. It is the point on the locus nearest to A and B, and this point suggests a line bisecting AB at right angles as the required locus.

To sum up, proceed as follows in attempting to solve a Rider:

I. Draw a careful figure.

II. Write down what is assumed and what is required.

III. Try to recall any similar construction or proof in Euclid.

IV. Take the Rider for granted, and try what can be deduced from it.

V. Draw several cases or figures for the Rider, and especially notice any critical cases of the figure.

MISCELLANEOUS RIDERS ON PROPOSITIONS 1—26. 1. The straight line which bisects the vertical angle of an isosceles triangle also bisects the base.

2. The straight line which bisects the vertical angle of an isosceles triangle is perpendicular to the base.

3. If the opposite sides of a quadrilateral figure are equal, show that the opposite angles are equal.

4. Two isosceles triangles are on the same base and on opposite sides of it; show that the straight line joining their vertices bisects the vertical angles.

5. The bisectors of the adjacent angles, which one straight line makes with another straight line on one side of it, are at right angles to one another.

6. The bisectors of two vertically opposite angles are in one straight line.

7. From a given point outside a given straight line only one perpendicular can be drawn to that line.

8. The perpendicular is the shortest line that can be drawn from a given point to a given straight line ; and of any other straight lines, one which is nearer to the perpendicular is always less than one more remote ; and from the same point two, and only two, equal straight lines can be drawn from the given point to the given straight line, one on each side of the perpendicular.

9. The difference of any two sides of a triangle is less than the third side. 10. The perimeter of a quadrilateral figure is greater than the sum of the diagonals, but less than twice the sum of the diagonals.

11. In an isosceles triangle the perpendiculars drawn from each of the equal angles to the opposite sides are equal.

12. Prove that if the bisectors of the angles at the base of a triangle meet at a point / within the triangle, the point I is equidistant from the three sides of the triangle.

13. Prove that if the exterior angles DBC, ECB of the triangle ABC made by producing AB and AC to D and E, be bisected by the straight lines BO and Co meeting in 0, the point 0 is equidistant from AD, BC, AE.

14. Find the locus of all points equidistant from two given points.

15. Find the locus of all points which are equidistant from two given straight lines which cut one another.

16. Find a point which is equidistant from three given straight lines. When is this impossible ?

17. On a given finite straight line as base, show how to construct a triangle having one of the angles at the base equal to a given angle, and the sum of the sides equal to a given straight line.

18. Given the base of a triangle, one of the angles at the base and the difference of the sides ; show how to construct the triangle.

19. Show how to construct a rectilineal figure equal in all respects to ą given rectilineal figure.

PART II.

PARALLEL STRAIGHT LINES AND PARALLELOGRAMS.

BEFORE commencing this Part of Euclid the Definitions of Parallel straight lines and of a Parallelogram, and the 12th Axiom should be learnt. If a straight line EF fall on

E two straight lines AB and CD, it will make eight angles, four A

-B angles being outside the lines AB and CD, and four angles

C

D within the lines AB and CD. The

H angles outside AB and CD are

F called exterior angles; and the angles within AB and CD are called interior angles.

The angles AGH and GHD are called alternate angles. The angles BGH and GHC are also alternate angles.

Euclid's 12th Axiom asserts that if the angles BGH, GHD are together less than two right angles, the lines AB and CD will meet towards B and D.

By Prop. 13 the angles AGH, BGH are together equal to two right angles; and also the angles GHD, GHC are together equal to two right angles. Therefore the four interior angles AGH, BGH, GHD, GHC are together equal to four right angles. If therefore the two angles BGH, GHD are together less than two right angles, the two angles AGH, GHC will be together greater than two right angles. According to the 12th Axiom therefore the lines AB and CD will meet towards B and D when the interior angles AGH, GHC are greater than the angles BGH, GHD. But if the angles BGH, GHD are together just equal to two right angles, the angles AGH, GHC will also be together equal to two right angles ; and the angles BGH, GHD will be together equal to the angles AGH, GHC. In this case we might expect the lines AB and CD not to meet, either towards B, D or A, C; in other words, AB and CD should be parallel. They are proved parallel by Euclid in Prop. 28.

PROP. 27.-- Theorem.-If a straight line, falling on two other straight lines in the same plane, make the alternate angles equal to one another, these two straight lines shall be parallel,

Let the straight line EF, falling on the two straight lines AB, CD in the same plane, make the angle AGH equal to the alternate angle GHD;

then AB shall be parallel to CD.

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Proof.—If AB and CD be not parallel, they will meet, if produced,

either towards B and D or towards A and C. If possible, let AB and CD, when produced, meet towards B and D

in the point K. Then KGH is a triangle, having the side KG produced to A, therefore the exterior angle AGH is greater than the interior opposite angle GHD.

Prop. 16 But the angle AGH is equal to the angle GHD, Нур. . therefore the angle AGH is both equal to and greater than the

angle GHD, which is impossible. Therefore AB and CD cannot meet, when produced, towards B and D. Similarly it may be proved that AB and CD cannot meet,

when produced, towards A and C.

Therefore AB is parallel to CD. Q. E. D.

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