EXERCISE XXVIII.

(For 1-13, refer to Diagram I.; for 14-19, to Diagram IV.)

1. Show, by I. 18, that the angle OCR is greater than the angle ORC. 2. Also, by I. 19, that in the triangle ERS the side ER must be greater than ES.

3. The angle EFT is obtuse; prove ET greater than either EF or FT. 4. Show that OG and GT together are greater than OP.

5. ER and RO are greater than ED and DO.

6. Which is the greater angle, EDO or ERO? Why?

7. Also EDC or ERC? Give the reason.

8. Show that EO, OT are equal to CO, OR, each to each.

If it be given that EOT is greater than COR, what is the inference with regard to CR and ET?

9. Given angle AOC greater than DOE, prove AC greater than DE. 10. Given angle POQ greater than QOR, prove PQ greater than QR. 11. Given angle EOF greater than GOF, prove ET greater than GT. 12. If CD be less than DE, show that the angle COD is less than DOE. 13. If ET be greater than ER, show that angle EOT is greater than EOR. 14. Find a straight line equal in length to AO and OC together. 15. Prove that AD is equal to EO and OC together.

16. Show that the angle OAE is less than AEC.

17. If BDF be equilateral, prove that the angle AFD is greater than the angle BDF.

18. If AFD be a right angle, show that DF is less than 40 and OC together.

19. The sum of the sides of the figure ABCDEF is greater than the sum of the sides of the triangle ACE.

EXAMINATION XXI.

1. Give the hypothesis and thesis of Prop. XVIII. and also of XIX. ; and state the relation of the one proposition to the other. 2. What are the respective methods of proof of XVIII. and of XIX. ? 3. How many false hypotheses are tested in the latter? State these, and also the false conclusions deduced from them. 4. What proposition may be enunciated thus: If two triangles are upon one base, and have the vertex of one within the other, show that the two sides of the inner triangle are less than those of the outer, but the inner vertical angle is greater than the outer one. 5. Where in the previous propositions has it been proved that the sides of the inner triangle cannot be respectively equal to the sides of the outer? 6. In Prop. XXII. count the number of printed lines devoted to the general enunciation, particular enunciation, construction, and proof respectively. 7. Mention a previous proposition which is a

particular case of Prop. XXII. 8. What is a Lemma? State an ex ample of a lemma among the propositions of Euclid which have been already given. 9. In the figure of XXIV., DG is taken equal to DF, and not less than DE. What is thus ensured with regard to the point F in the figure?

SECTION V.-PROPOSITION XXVI.

Third and Fourth Cases of Equality of Two Triangles. EXERCISE XXIX.

1. In the triangle PQR, which side is adjacent to

both the angles Q, R?

2. Given that angles Q, R are equal to angles Y, Z, each to each, and the sides adjacent to those angles are also equal; write out these given parts of the two triangles as three pairs of equals.

3. Write out the remaining sides in pairs, the two in each pair being opposite to equal angles.

R

4. Given the angles Q, R equal to Y, Z, each to each, and the sides also equal which are opposite the first pair of equal angles; write these parts of the two triangles in three pairs of equals.

5. Write out the pair of third angles.

6. Given angles P, R equal to X, Z, each to each, and the sides also equal which are opposite the

Ꮓ

second pair of equal angles; write these in three pairs of equals.

7. Given in PQR, XYZ that PQ, QR, and angle Q, are respectively equal

to XY, YZ, and angle Y; show that angles R and Z are equal.

8. In the annexed figure, show that the

angle V is less than the angle STU.

9. Given that SU, UV, and the angle U, are equal to XY, YZ, and the angle Y; show that the angle STU is greater than the angle Z.

10. Given that SU, UV, and the angle U, are equal to PQ, QR, and the angle Q; show that

the angle P is greater than the angle UST.

T

11. Given ST, TU, US respectively equal to PR, RQ, QP; determine which is the greater of the angles R and V.

65. PROPOSITION XXVI.-THEOREM.

If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, namely, either the sides adjacent to the equal angles or the sides opposite to one pair of equal angles; then shall their other sides be equal, each to each, and also their third angles.

Let ABC, DEF be two triangles which have the angles ABC, BCA equal to the angles DEF, EFD, each to each, namely, ABC to DEF, and BCA to EFD;

also one side equal to one side.

First, let those sides be equal which are adjacent to the equal angles, namely, BC to EF.

Then the other sides shall be equal, each to each,

namely, AB to DE, and AC to DF,

and the third angle BAC to the third angle EDF. For, if AB be not equal to DE,

one of them must be greater than the other. Let, if possible, AB be greater than DE; make BG equal to ED (I. 3), and join GC.

Then, in the two triangles GBC, DEF,

because GB is equal to DE, and BC to EF;

and the angle GBC is equal to the angle DEF;

(Hyp.)

therefore those angles are equal which are opposite the equal

sides GB, DE,

that is, the angle GCB is equal to the angle DFE.

But DFE is equal to ACB.

Wherefore, also, GCB is equal to ACB,

the less equal to the greater, which is impossible; therefore AB is not unequal to DE,

that is, AB is equal to DE.

(I. 4)

Hence, in the triangles ABC, DEF, because AB is equal to DE, and BC to EF, and the angle ABC to the angle DEF,

therefore the base AC is equal to the base DF, and the third angle BAC to the third angle EDF.

(I. 4)

Secondly, let sides in each triangle be equal which are op

posite one pair of equal angles, namely, AB equal to DE.

Then, in this case likewise, the other sides shall be equal,

namely AC to DF, and BC to EF,

and also the third angle BAC to the third angle EDF.

For, if BC be not equal to EF,

one of them must be greater than the other. Let, if possible, BC be greater than EF, make BH equal to EF (I. 3), and join AH.

D

B

H C E

Then, in the two triangles ABH, DEF,

F

because AB is equal to DE, and BH to EF,
and the angle ABH to the angle DEF;
therefore the angles opposite to AB, DE are equal,

that is, the angle BHA is equal to the angle EFD.

But the angle EFD is equal to BCA ;

(Hyp.)

(I. 4)

(Hyp.)

therefore BHA is equal to BCA, which is impossible. (I.16)

Wherefore BC is not unequal to EF,

that is, BC is equal to EF.

Hence, in the triangles ABC, DEF,

because AB is equal to DE, and BC to EF, and the angle ABC is equal to the angle DEF, therefore the base AC is equal to the base DF, and the third angle BAC to the third angle EDF. Wherefore, if two triangles, &c.

(Hyp.)

(I. 4)

Q. E. D.

66. ON PROP. XXVI.

This proposition is double. It contains two of the cases of equality of triangles; the first of which may be separately enunciated thus:

If two triangles have two angles of the one equal to two angles of the other, each to each, and also the sides adjacent to these angles equal; they shall have all their other elements equal in pairs.

The enunciation of the second case of Prop. XXVI. after this manner will be obvious.

These two cases may each, like Props. IV. and VIII., be proved by superposition, as shown in the Appendix.

EXERCISE XXX.

1. In the triangles PQR, XYZ, given angle P equal to X, and Q equal to Y, and base QR equal to YZ; which part of Prop. XXVI. applies to this case? (Figure on page 73.)

Give the remaining pairs of elements which are equal by that proposition.

2. Given, in the same triangles, that angles P and Q are respectively equal to Y and Z; which pair of sides must be equal in order that the first part of XXVI. may be applicable?

Write out the resulting pairs of equals also.

Questions 3-8 refer to the annexed diagram, in
which it is given that UX, UY are equal.

3. If UV bisects the angle XUY, find two angles and a side in the triangle UVX which are respectively equal to two angles and a side of the triangle UVY. Let the side in each triangle be adjacent to the angles.

A

4. Also find two angles and a side in one triangle equal to similar parts in the other, but the sides being opposite a pair of equal angles.

5. Deduce from either case that UV bisects the base, and is perpendicular to it.

6. Next, given that UV is perpendicular to XY, find two angles and a side in one of the triangles respectively equal to two angles and a side in the other, so as to fulfil the conditions of the second part of XXVI.

7. Also find similar data, but taking a different pair of sides.

8. Infer from either set of data that the perpendicular bisects the base, and also the angle XUY.

9. If any angle be bisected by a straight line, any point in this bisector is equally distant from the two straight lines which contain the angle. (By the distance of a point from a line is meant the perpendicular from the point to the line.)