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1. The triangle ABC is equal to the triangle GEF (I. 38), because they are upon equal bases BC, EF, and between the same parallels BF, AG; but the triangle ABC is equal to the triangle DEF (hyp.); therefore

2. The triangle DEF is equal to the triangle GEF (Ax. 1), the greater triangle equal to the less, which is impossible ; therefore

3. AG is not parallel to BF. And in the same manner it can be demonstrated, that there is no other line drawn from A parallel to it but AD; therefore 4. AD is parallel to BF.

Wherefore, equal triangles upon, &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.

Let the parallelogram ABCD, and the triangle EBC be upon the same base BC, and between the same parallels BC, AE.

Then the parallelogram ABCD shall be double of the triangle EBC.

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Demonstration. Then

D

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B

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

2. The parallelogram ABCD is double of the triangle ABC, because the diameter AC bisects it (I. 34); wherefore also

3. ABCD is double of the triangle EBC.
Therefore, if a parallelogram and a triangle, &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.

Let ABC be the given triangle, and D the given rectilineal angle. 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.

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Construction. Bisect BC in E (I. 10), and join AE; at the point E, in the straight line EC, make the angle CEF equal to the angle D (I. 23); through C draw CG parallel to EF; and through A draw AFG parallel to BC (I. 31), meeting EF in F, and CG in Ğ.

Then the figure CEFG is a parallelogram (Def. A). Proof. Because the triangles ABE, AEC are on the equal bases BE, EC, and between the same parallels BC, AG; therefore

1. The triangle ABE is equal to the triangle AEC (I. 38), and

2. The triangle ABC is double of the triangle A EC; but

3. The parallelogram FECG is double of the triangle A EC

(I. 41), because they are upon the same base EC, and between the same parallels , AG; therefore

4. The parallelogram FECG is equal to the triangle ABC

(Ax. 6), and it has one of its angles CEF, equal to the given angle D.

Wherefore, 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 diameter of any parallelogram, are equal to one another.

Let ABCD be a parallelogram, of which the diameter is AC; and EH, GF the parallelograms about AC, that is, through which AC passes: also BK, KD the other parallelograms which make up the whole figure ABCD, which are therefore called the complements.

Then the complement BK shall be equal to the complement KD.

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Demonstration. Because ABCD is a parallelogram, and AC its diameter, therefore

1. The triangle ABC is equal to the triangle ADC (I. 34).

Again, because EKHA is a parallelogram, and AK its diameter, therefore

H

A

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2. The triangle AEK is equal to the triangle AHK (I. 34); and for the same reason,

3. The triangle KGC is equal to the triangle KFC. Wherefore

4. The two triangles A EK, KGC are equal to the two triangles

AHK, KFC (Ax. 2); but the whole triangle ABC is equal to the whole triangle ADC; therefore

5. The remaining complement BK, is equal to the remaining complement KD (Ax. 3).

Wherefore, 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.

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Let AB be the given straight line, and C the given triangle, and D the given rectilineal angle. It is required to apply to the straight line AB a parallelogram equal to the triangle C, and having an angle equal to the angle D.

Construction (I.) Make the parallelogram BEFG equal to the triangle C, and having the angle EBG equal to the angle D (I. 42), so that BE be in the same straight line with AB; produce FG to H; through A draw AH parallel to BG or EF (I. 31), and join HB. Proof (I.) Then because the straight line HF falls upon

the

parallels AH, EF, therefore

1. The angles AHF, HFE are together equal to two right

angles (I. 29); Wherefore

2. The angles BHF, AFE are less than two right angles

(Ax. 9, but straight lines which with another straight line make the two interior angles upon the same side less than two right angles, do meet if produced far enough (Ax. 12); therefore

3. HB and FE shall meet if produced. Construction (II.) Let HB and FE be produced and meet in K; through K draw ŹL parallel to EA or FH, and produce HA, GB to meet KL in the points L, M.

Then LB shall be the parallelogram required.

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

4. The complement LB is equal to the complement BF (I. 43), but the complement BF is equal to the triangle C (constr.); wherefore

5. LB is equal to the triangle C. And because the angle GBE is equal to the angle ABM (I. 15), and likewise to the angle D (constr.); therefore

6. The angle ABM is equal to the angle D (Ax. 1). Therefore to the given straight line AB the parallelogram LB has been applied, equal to the triangle C, and having the angle ABM equal to the given 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.

Let ABCD be the given rectilineal figure, and E the given recti

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