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Demonstration.-1. The triangle ABC is equal to the triangle GEF, because they are upon equal bases BC, EF, and are assumed to be between the same parallels BF, AG. (I. 38.)

2. But the triangle ABC is equal to the triangle DEF,

3. Therefore the triangle DEF is equal to the triangle GEF (ax. 1); the greater equal B

C E to the less, which is impossible.

4. Therefore AG is not parallel to BF.

5. In the same manner, it can be demonstrated that no line, passing through A, can be parallel to BF, except AD;

6. Therefore AD is parallel to BF.
Conclusion.—Therefore, equal triangles, &c. Q. E. D.

PROPOSITION 41.-THEOREM.

If a parallelogram and a triangle be upon the same base, and between the same parallels, the parallelogram shall be double of the triangle.

(References—Prop. I. 34, 37; ax. 1.) Hypothesis.—Let the parallelogram ABCD, and the triangle EBC be upon the same base BC, and between the same parallels BC, AÊ. Sequence.—The parallelogram ABCD

DE shall be double of the triangle EBC.

Construction.—Join AC.

Demonstration. — 1. The triangle ABC is equal to the triangle EBC, because they are upon the same base BC, and between the same parallels BC, AE. (I. 37.).

2. But the parallelogram ABCD is double of the triangle ABC, because the diagonal AC bisects it. (I. 34.)

3. Therefore the parallelogram ABCD is also double of the triangle EBC. (ax. 1.)

Conclusion. Therefore, if a parallelogram, &c. Q. E. D.

PROPOSITION 42.- PROBLEM. To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.

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(References—Prop. I. 10, 23, 31, 38, 41; ax. 6; def. 35.) Given.—Let ABC be the given triangle, and D the given rectilineal angle.

Sought.-It is required to describe a parallelogram that shall be equal to the given triangle ABC, and have one of its angles equal to D.

Construction.-1. Bisect BC in E (I. 10), and join AE.

2. At the point E, in the straight line CE, make the angle CEF equal to D. (I. 23.)

3. Through' A draw AFG parallel to EC. (1. 31.)

4. Through Ć draw CG parallel to B E EF. (I. 31.)

Then FECG is the parallelogram required.

Proof.-1. Because BE is equal to EC (const.), the triangle ABE is equal to the triangle AEC, since they are upon equal bases and between the same parallels. (I. 38.)

2. Therefore the triangle ABC is double of the triangle AEC.

3. But the parallelogram FECG is also double of the triangle AEC, because they are upon the same base, and between the same parallels. (1. 41.)

4. Therefore the parallelogram FECG is equal to the triangle ABC; (ax. 6.)

5. And it has one of its angles CEF equal to the given angle D. (const.)

Conclusion. Therefore, a parallelogram FECG has been described equal to the given triangle ABC, and having one of its angles CEF equal to the given angle D. Q. E. F.

PROPOSITION 43.-THEOREM, The complements of the parallelograms which are about the diagonal of any parallelogram, are equal to one another.

(References—Prop. I. 34; ax. 2, 3.) Hypothesis.--Let ABCD be a parallelogram, of which the diagonal is AC; and EH, GF, parallelograms about AC, that is, through which AC passes; and BK, KD the other parallelograms, which make up the whole figure ABCD, and are therefore called the complements.

Sequence. The complement BK shall be equal to the complement KD.

Demonstration.-1. Because ABCD is a parallelogram, and AC its diagonal, the triangle ABC is equal to the triangle ADC. (I. 34.)

2. Again, because AEKH is a parallelogram, and AK its diagonal, the triangle AEK is equal to the triangle AHK. (I. 34.)

3. For the like reason the triangle KGC is equal to the triangle KFC. 4. Therefore, because the triangle

Роше и газов от AEK is equal to the triangle AHK, and the triangle KGC to KFC;

5. The triangles AEK, KGC, are equal to the triangles AHK, KFC. (ax. 2.)

6. But the whole triangle ABC was proved equal to the whole triangle ADC.

7. Therefore the remaining complement BK is equal to the remaining complement KD. (ax. 3.)

Conclusion.—Therefore, the complements, &c. Q. E. D.

PROPOSITION 44.—PROBLEM. To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.

(References, Prop. I. 15, 29, 31, 42, 43; ax. 1, 9, 12.) Given. Let AB be the given straight line, C the given triangle, and D the given angle.

Sought.--It is required to apply to the straight line AB, a parallelogram equal to the triangle C, and having an angle equal to D.

Construction.—(I.) 1. Make the parallelogram BEFG equal to the triangle C, and having the angle EBG equal to the angle D. (1. 42.)

2. And let the parallelogram BEFG be made so that BE may be in the same straight line with AB.

3. Produce FG to H.
4. Through A draw AH parallel to BG or EF. (I. 31.)
5. Join HB.

Proof. — (I.) 1. Because the straight line HF falls on the parallels AH, EF, the angles AHF, HFE, are together equal to two right angles. (I. 29.)

2. Therefore the angles BHF, ÁFE, are together less than

two right angles. (ax. 9.) But straight lines which with another straight line make the interior angles on the same side together less than two right angles, will meet on that side, if produced far enough. (ax. 12.)

3. Therefore HB and FE shall meet if produced.

Construction.—(II.) 1. Produce HB and FE towards BE, and let them meet in K.

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2. Through K draw KL parallel to EA or FH. (I. 31.)
3. Produce HA, GB, to the points L, M.
Then LB shall be the parallelogram required.

Proof.—(II.) 1. Because HLKF is a parallelogram, of which the diagonal is HK; and AG, ME, are the parallelograms about HK; and LB, BF are the complements;

2. Therefore the complement LB is equal to the complement BF. (I. 43.)

3. But BF is equal to the triangle C. (const.)
4. Therefore LB is equal to the triangle C. (ax. 1.)

5. And because the angle GBE is equal to the angle ABM (I. 15), and likewise to the angle D, (const.)

6. Therefore the angle ABM is equal to the angle D. (ax. 1.)

Conclusion. Therefore, the parallelogram LB is applied to the straight line AB, and is equal to the triangle C, and has the angle ABM equal to the angle D. Q. E. F.

PROPOSITION 45.—PROBLEM.

To describe a parallelogram equal to a given rectilineal figure,

and having an angle equal to a given rectilineal angle. (References—Prop. I. 14, 29, 30, 42, 44; ax. 1, 2; def. 35.)

Given.—Let ABCD be the given rectilineal figure, and E the given rectilineal angle.

Sought. It is required to describe a parallelogram equal to ABCD, and having an angle equal to E.

Construction.-1. Join DB.

2. Describe the parallelogram FH equal to the triangle ADB, and having the angle FKH equal to the angle E. (1. 42.)

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3. To the straight line GH apply the parallelogram GM equal to the triangle DBC, and having the angle GHM equal to the angle E. (I. 44.)

Then the figure FKML shall be the parallelogram required.

Proof.–1. Because the angle E is equal to each of the angles FKH, GHM, (const.)

2. Therefore the angle FKH is equal to the angle GHM. (ax. 1.)

3. Add to each of these equals the angle KHG;

4. Therefore the angles FKH, KHG, are equal to the angles KHG, GHM. (ax. 2.)

5. But FKH, KHG, are equal to two right angles. (I. 29.)

6. Therefore also KÉG, GÁM, are equal to two right angles. (ax. 1.)

7. And because at the point H in the straight line GH, the two straight lines KH, HM, on the opposite sides of it, make the adjacent angles together equal to two right angles;

8. Therefore KH is in the same straight line with HM. (I. 14.)

9. And because the straight line HG meets the parallels KM, FG, the alternate angles MHG, HGF, are equal. (I. 29.)

10. Add to each of these equals the angle HGL;

11. Therefore the angles MHG, HGL, are equal to the angles HGF, HGL. (ax. 2.)

12. But the angles MHG, HGL, are equal to two right angles; (I. 29.)

13. Therefore also the angles HGF, HGL, are equal to two right angles.

14. And therefore FG is in the same straight line with GL. (I. 14.)

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