Sidebilder
PDF
ePub

1. The two sides AB, AC, equal to the two DE, DF, each to each, namely, AB to DE, and AC to DF,

2. But the angle BAC greater than the angle EDF. Sequence.--The base BC shall be greater than the base EF.

Construction.-1. Let the side DF of the triangle DEF be greater than its side DE.

2. Then at the point D, in the straight line ED, make the angle EDG equal to the angle BAC. (I. 23.)

3. Make DG equal to AC or DF. (I. 3.)* !

4. Join EG, GF.

Demonstration. — 1. B Because AB is equal to DE (hyp.), and AC to DG (const.), the two sides BA, AC, are equal to the two ED, DG, each to each.

2. And the angle BAC is equal to the angle EDG. (const.) 3. Therefore the base BC is equal to the base EG. (I. 4.)

4. And because DG is equal to DF (const.), the angle DFG is equal to the angle DGF. (I. 5.)

5. But the angle DGF is greater than the angle EGF. (ax. 9.) 6. Therefore the angle DFG is greater than the angle EGF.

7. Much more then is the angle EFG greater than the angle EGF.

8. And because the angle EFG of the triangle EFG is greater than its angle EGF, and that the greater angle is subtended by the greater side,

9. Therefore the side EG is greater than the side EF. (I. 19.) 10. But EG was proved equal to BC; 11. Therefore BC is greater than EF. Conclusion.--Therefore, if two triangles, &c. Q. E. D.

PROPOSITION 25.—THEOREM. If two triangles have two sides of the one equal to two sides of the other, each to each, but the base of the one greater than the base off the other, the angle contained by the sides of that which has the greater buse, shall be greater than the angle contained by the sides equal to thein, of the other.

(References—Prop. I. 4, 24.) Hypothesis. — Let ABC, DEF, be two triangles, which bave

1. The two sides AB, AC, equal to the two sides DE DF, each to each; namely, AB to DĒ, and AC to DF.

2. But the base BC greater than the base EF.

Sequence.—The angle BAC shall be greater than the angle EDF.

Demonstration.-1. For if the angle BAC be not greater than the angle EDF, it must either be equal to it or less.

2. But the angle BAC is not equal to the angle EDF, for then the base BC would be equal to the base EF (I. 4), but it is not. (hyp.)

3. Therefore the angle BAC is not equal to the angle EDF. B

4. Neither is the angle BAC less than the angle EDF, for then the base BC would be less than the base EF (I. 24), but it is not. (hyp.)

5. Therefore the angle BAC is not less than the angle EDF.

6. And it has been proved that the angle BAC is not equal to the angle EDF.

7. Therefore the angle BAC is greater than the angle EDF. Conclusion.—Therefore, if two triangles, &c. Q. E. D.

PROPOSITION 26.-THEOREM.

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

If two angles and a side in one triangle be respectively equal to two angles and a corresponding side in another triangle, the triangles shall be egual in every respect.

(References—Prop. I. 3, 4, 16; ax. 1.) Hypothesis.—Let ABC, DEF, be two triangles, which have

1. The angles ABC, BCA, equal to the angles DEF, EFD, each to each; namely, ABC to DEF, and BCĂ to EFD.

2. Also one side equal to one side.

Case I.–First, let the sides adjacent to the equal angles in each be equal; namely, BC to EF.

Sequence. Then shall the side AB be equal to DE, the side AC to DF, and the angle BAC to the angle EDF.

Hypothesis.-(II.) For if AB be not equal to DE, one of them must be greater than the other. Let AB be the A greater of the two.

Construction. — Make Gh BG equal to DE (I. 3), and join GC.

Demonstration.-1. Because BG is assumed to be equal to DE (const.), and

C E BC is equal to EF (hyp.), the two sides GB, BC, are equal to the two sides DE, EF, each to each.

2. And the angle GBC is equal to the angle DEF, (hyp.) 3. Therefore the base GC is equal to the base DF, (I. 4.) 4. And the triangle GBC to the triangle DEF, (I. 4.)

5. And the other angles to the other angles, each to each, to which the equal sides are opposite.

6. Therefore the angle GCB is equal to the angle DFE. (I. 4.) 7. But the angle DFE is equal to the angle BCA. (hyp.)

8. Therefore the angle GCB is equal to the angle BCA (ax. 1), the less to the greater, which is impossible.

9. Therefore AB is not unequal to DE, that is, it is equal to it; and BC is equal to EF. (hyp.)

10. Therefore the two sides AB, BC, are equal to the two sides DE, EF, each to each.

11. And the angle ABC is equal to the angle DEF, Cyp.) 12. Therefore the base AC is equal to the base DF, (I. 4.) 13. And the third angle BAC to the third angle EDF. (I. 4.)

Case II.-(Hypothesis.) Next, let the sides which are opposite to the equal angles in each triangle be equal to one another; namely, AB equal to DE.

Sequence.—Likewise in this case the other sides shall be equal, AC to DF, and BC to EF; and also the angle BAC to the angle EDF.

Hypothesis.--(II.) For if BC be not equal to EF, one of them must be greater than the other. Let BC be the greater of the two.

B H C B
Construction. - Make
BH equal to EF (I. 3), and join AH.

Demonstration.-l. Because BH is assumed to be equal to

EF (const.), and AB is equal to DE (hyp.), the two sides AB, BH, are equal to the two sides DE, EF, each to each.

2. And the angle ABH is equal to the angle DEF. (hyp.) 3. Therefore the base AH is equal to the base DF, I. 4.) 4. And the triangle ABH to the triangle DEF, (1. 4.)

5. And the other angles to the other angles, each to each, to which the equal sides are opposite.

6. Therefore the angle BHA is equal to the angle EFD. (1.4) 7. But the angle EFD is equal to the angle BCA. (hyp.)

8. Therefore the angle BHÀ is also equal to the angle BCA. (ax. 1.)

9. That is, the exterior angle BHA of the triangle AHC, is equal to its interior and opposite angle BCA, which is impossible. (I. 16.)

10. Therefore BC is not unequal to EF, that is, it is equal to it; and AB is equal to DE. (hyp.)

11. Therefore the two sides AB, BC, are equal to the two sides DE, EF, each to each.

12. And the angle ABC is equal to the angle DEF. (hyp.) 13. Therefore the base AC is equal to the base DF, I. 4.)

14. And the third angle BAC is equal to the third angle EDF. (I. 4.)

Conclusion. Therefore, if two triangles, &c. Q. E. D.

PROPOSITION 27.—THEOREM. If a straight line falling upon two other straight lines make the alternate angles equal to one another, these two straight lines shall be parallel.

(References—Prop. I. 16; def. 34.) Hypothesis.-Let the straight line EF, which falls upon the two straight lines AB, CD, make the alternate angles AEF, EFD, equal to one another.

Sequence.—AB shall be parallel to CD.

Hypothesis.-(II.) For if AB and CD be not parallel, they will meet if produced, either towards

C /
B, D, or towards Á, C. Let them
be produced, and meet towards
B, D, in the point G.

Demonstration.-). Then GEF is a triangle, and its exterior angle AEF is greater than the interior and opposite angle EFG. (I. 16.)

2. But the angle AEF is also equal to EFG (hyp.), which is impossible.

3. Therefore AB and CD, being produced, do not meet towards B, D.

4. In like manner it may be shown that they do not meet towards A, C.

5. But those straight lines which being produced ever so far both ways do not meet, are parallel. (def. 34.)

6. Therefore AB is parallel to CD. Conclusion.---Therefore, if a straight line, &c. Q. E. D.

PROPOSITION 28.--THEOREM.

If a straight line falling upon two other straight lines make the exterior angle equal to the interior and opposite upon the same side of the line, or make the interior angles upon the same side together equal to two right angles, the two straight lines shall be parallel to one another.

(References—Prop. I. 13, 15, 27; ax. 1, 3.) Hypothesis.—Let the straight line EF, which falls upon the two straight lines AB, CD, make

1. The exterior angle EGB equal to the interior and opposite angle GHD, upon the same side;

2. Or make the interior angles on the same side, the angles BGH, GHD, together equal to two right A angles.

Sequence.—AB shall be parallel to CD.

Demonstration. — (I.) 1. Because the angle EGB is equal to the angle GHD, (hyp.)

2. And the angle EGB is equal to the angle AGH, (I. 15.)

3. Therefore the angle AGH is equal to the angle GHD (ax. 1), and these angles are alternate.“

4. Therefore AB is parallel to CD. (I. 27.)

Demonstration.—(II.) 1. Again, because the angles BGE, GHD, are equal to two right angles, (hyp.)

2. And the angles AGH, BGH, are also equal to two right angles, (I. 13.)

3. Therefore the angles AGH, BGH, are equal to the angles BGH, GHD. (ax. 1.)

4. Take away the common angle BGH.

« ForrigeFortsett »