result that we know to be true or false, in order to make this result the basis for the proof or disproof of the assumed theorem.

The question arises, Is it always necessary to begin the solution of a given exercise by this analytic process? Certainly not, as regards theorems, at least, though it is usually necessary with problems. Most of the theorems given as exercises are of such a simple character that, once they are clearly understood and the diagram prepared, their dependence upon some known theorem is obvious, and we can proceed at once to the synthetic proof. At the same time it may be recommended, as an advantageous training for the attack of more difficult exercises, to take, in every case, as many steps in the analysis as will make the synthetic basis obvious.

Thus in Ex. 112, though it is evident that the theorem is merely a special case of Prop. IV., yet, designating the prolongation of BA as AD, we may reason thus: If CAB, CAD are both right angles, BAD is a straight line. But BAD is a straight line (Hyp.); etc.

=

In Ex. 120, if ABC be the triangle, and the bisectors meet in O, it is seen at once that since ≤ OBC = ≤ OCB, being halves of equal angles, it follows as an immediate consequence that OC OB. Yet we may reason thus: If OC = OB, then ≤OBC= 40CB; but these angles are equal, being halves of equal angles; etc.

In Ex. 147 the dependence is not obvious, so we proceed as follows:

DCE or

≤ACE =

(We make this addition to both members of the equality because, supposing BC produced to E, we perceive that ZDBC or ABC+2D, while LACE = 2 ABC + 2A). Examination of the diagram showing us that this equality is true,

we proceed to the synthesis:

Since ≤ DCE = 2 DBC or } ≤ ABC + 2D,

(122)*

and DCE or

LACE =

ZABC + ZA, S

(122)

ZABC + 2D

=

ZABC +

ZA; .. ZD=

24.

Thus far we have needed no addition to the origi

nal diagram.

In Ex. 149 let PD, PE be the distances from a point P in the base to AB, AC respectively. Draw the altitude BF to AC.

=

If PD+ PE BF, as the theorem asserts, then by cutting off a part of BF equal to PE, as FG, we obtain GB=PD. We evidently can obtain the point G by

*The reference is to Art. 122 of the Geometry.

drawing PG parallel to AC, since then PF will be a parallelogram; and, producing PG to meet AB in H, we have an isosceles ▲ HBP, since HPB = LC = LB. Then as PD is perpendicular to AB (Hyp.), and BG is perpendicular to PH, being perpendicular to its parallel AC, we have the altitudes to the arms of isosceles ▲HPB equal. Now this is true by Ex. 119; hence the synthesis:

Draw PH || to AC, to meet the altitude BF in G, and AB in H. Since HPB = ZC (112) = ZB, ▲ HPB is isosceles; and as PD is perpendicular to HB (Hyp.), and BG is perpendicular to PH, being perpendicular to AC, PD = BG (Ex. 119), and PF being a parallelogram by construction, PE = GF; ... PD + PE = BG + GF = BF. Special attention should be paid to the foregoing analysis, as it affords an example of what almost always occurs, unless the analysis is of the simplest character; that is, the mingling with it of a certain amount of synthesis. Thus the second sentence of the analysis, that beginning, "We evidently," is wholly synthetic in character, being quite independent of the assumed theorem. This combination of the two methods is that most practically useful. Whenever we make an addition to the original diagram, such as drawing a parallel to some line, we have to develop the relations of such lines synthetically in order to help out our analysis; while, again, in our synthetical reasoning, if we come to a result that seems as if it ought to be true, though we are not able to say that it is so, we may help ourselves by inquiring, If this were true, what consequences would result from it ?

In conclusion, it may be remarked that the analytic method is best suited to exercises in which special results or equalities are to be established, as in the foregoing examples. It is much less suited to exercises in which general results or inequalities are to be established. Thus if, in Ex. 142, we let AM, BN, CP, be the medians from A, B, C, resp., we will find that to assume the inequality AM+BN+CP<AB+AC+BC, leads to the desired result by a somewhat roundabout way. Slight study of the diagram will, on the other hand, suggest the synthesis AM<}(AB+AC), BN<\(AB+BC), CP<}(AC+BC); (Ex. 138) .. AM+ BN + CP<{(2 AB + 2 AC + 2 BC)< AB+ AC + BC. Similarly, for the second part of the theorem,

B

M

AM>{(AB + AC – BC), BN>{(AB+ BC — AC),

CP>}(AC+ BC − AB);

.'. AM+ BN + CP>}(AB+ AC+BC).

Little need be said about the third general direction, as it may reasonably be assumed that only true theorems will be given as exercises, yet it finds an important application whenever, as before adverted to, we arrive at some suggested relation which we do not see how to prove. We assume the thing to be true, and trace what consequences would result from it. If we are thus led to a contradiction of some established truth, or to something inconsistent with the data of the given question, we know that the suggested relation cannot be true, and reject it accordingly.

QUESTIONS, p. 17.

(Only slight hints are given of the answers to most of these questions.)

1. AB = AC +CB; AC = AB - CB. 2. A straight line. 3. A right angle. 4. A straight line. 5. A curved line. 6. Plane surfaces, which are also presented by the surface of still water, of a pane of glass, etc. 7. By applying the straightedge to the given surface and twisting it about in order to ascertain whether the edge always coincides with the given surface. Mechanics often oil the surface that is to be tested in this way. Try this on a slate. 8. The surface of a stovepipe is not a plane, because straight lines can be drawn on it only in a certain direction, and not between any two points. Illustrate the same thing on a conical surface. 9. Only a curve can be drawn on the surface of an eggshell. Illustrate the same thing on the surface of a round pebble or apple. 10. The angles are AOB, BOC, COD, AOC, AOD, BOD. Of these, AOB and BOC are adjacent, as are also AOC and COD, BOC and COD. AOC is the sum of two angles, AOB and BOC, while AOD is the sum of AOB, BOC, and COD. 11. Yes; in the preceding figure AOB and AOC have the common vertex O and the common side OA, yet are not adjacent. 12. An acute angle is less than, a right angle is equal to, an obtuse angle is greater than, its supplement.

QUESTIONS, p. 21.

D

C

1. See p. 20. 2. We first draw a straight line nearly as long as the straightedge, then shift the latter so as to coincide with only a part of the line just drawn. The part of the line now drawn will have the same direction as the first part (15). 3. We open the dividers or compasses so that the points coincide with the extremities of CD. If we then place one of the points on an extremity of AB, the other point will mark off a distance equal to CD (20). 4. BE is laid off

equal to CD on AB produced, as shown in the preceding answer. Then AE = AB + BE = AB + CD and BE AE – AB. = 5. An obtuse angle. 6. An acute angle. 7. The angles are BAC, CAD, and BAD. 8. Confining ourselves to one side of BAE, we have the six angles, BAC, CAD, DAE, BAD, CAE, BAE. 9. The sum of the angles BAC and CAD is equal to the angle BAD. 10. The difference of the angles BAD and BAC is equal to the angle CAD. 11. The angle BAD is greater than the angle CAD. 12. The angle BAC is less than the angle BAD. 13. AB+BC=AC. 14. AC—BC=AB. 15. AC> BC. 16. AB < AC. 17. Z BAD + 2BAC is equal to a right angle. 18. BAD - ≤BAC : = L CAD.

EXERCISES, pp. 24-68.

1. Let A denote the angle, and

Z A + } ¿ A = a rt. ≤ ; 2. Let A denote the angle, and

A its complement. Then
LA= rt. 2 = st. Z.

its supplement. Then

ZA + ZA = a st. Z; ... ZA = 3 st. Z

A

B

= rt. Z.

3. Let AE, AF be the bisectors of the adjacent complementary angles CAB, CAD.

i.e., ≤ CAE + ≤ CAF = { rt. 2.

4. LEAD is complement of ZEAC, and ≤ EAC of ▲ EAD.

5. ZEAB is supplement of LEAD, LEAD of ZEAB, and ▲ CAD of Z CAB, while ▲ CAB + LEAD is the supplement of CAE.

6. For EAb being a st. Z and CAB a rt. 2, BAb + CAE = a rt. 2. 7. If / BAC+ BAC a st. 2, then BAC = st. <= 1 rt. 2, and ≤ CAD or } ≤ BAC = } st. ≤ = 2 rt. 2.

8. For ZAOB + 2 AOD = ≤ DOC + ≤ BOC = a st. Z; that is, AOB and AOD are supplementary, as are also DOC and BOC. 9. In order that BA may be in a straight line with AD, BAD must be a st. ; or, by the conditions of the question, Z BAC+ rt. Z must equal two rt. ; i.e., ▲ BAC = 11 rt. 2.

11. The angles A, A' are the supplements of the angles formed by producing A, A' in both directions, and the angles thus formed are equal (44).

12. The exterior angles at C are equal, being the supplements of angles ACB, A'CB', which have been proved equal.

13. For then it would have two perpendiculars drawn from the same point to the same line, which is impossible (51).

14. For these exterior angles are the supplements of equal angles. 15. They cannot be right angles, by Ex. 13, much less can they be obtuse. The third angle may be acute, right, or obtuse.

16. Since the interior angles are unequal, the exterior angles that are respectively supplementary to them must be unequal.

17. For each of the two exterior angles thus formed is a right angle. 18. For each of the two exterior angles thus formed, being the supplement of an obtuse angle, must be acute.

19. For A will coincide with A', while O retains its position.

20. Since AOC, A'OC can be made to coincide, they are right angles (30). The bisection of A and A', and that of BC, can be proved by an anticipation of the demonstration of Prop. VIII., by making ZACA' coincide with ZABA'; etc.

=

21. For this we assume that OB OC has been proved, as in Ex. 20. 22. AO A'O, OC is common, and ZAOC = ZA'OC; etc.

=

23. In the first case AC would fall without ▲ A'; in the second, AC would fall within A'.

24. (1) The angles opposite unequal sides cannot be equal, since then the sides opposite them would be equal. (2) The sides opposite unequal angles cannot be equal, since then the angles opposite them would be equal.

25. For if it had two equal angles, it would have two equal sides, and would not be scalene.

=

ZC'A'A, ZB'AA' = Z B'A'A;

... ¿ C'AB' = Z C'A'B' (Ax. 3); etc.

29. Since the three sides of the one are respectively equal to the three sides of the other, the triangles are equal (69) and must coincide if they are on the same side of the common base.

30. Place the triangles as in the left-hand figure of Art. 74, and join the vertices. Then the halves of

the equal vertical angles being equal (74), the sides

opposite those equal angles are equal; hence the triangles are equal