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PROP. XXXVIII.-THEOREM.

Triangles upon equal bases and between the same parallels, are equal to one another.

Let the triangles ABC, DEF, be upon equal bases BC, EF, and between the same parallels BF, AD. The triangle ABC is equal to the angle DEF.

H

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D

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Produce AD both ways to the points G, H, and through B draw BG parallel (I. 31.) to CA, and through F draw FH parallel to ED. Then (I. 34. Def.)

1. Each of the figures GBCA, DEFH, is a parallelogram; and (I. 36.)

2. GBCA, DEFH, are equal to one another, because they are upon equal bases BC, EF, and between the same parallels BF, GH; and (I. 34.)

3. The triangle ABC is the half of the parallelogram GBCA, because the diameter AB bisects it; and

4. The triangle DEF is the half of the parallelogram DEH, because the diameter DF bisects it. But the halves of equal things are equal (Ax. 7.); therefore

5. T'he triangle ABC is equal to the triangle DEF. Wherefore, triangles, &c.

Q.E.D.

PROP. XXXIX.-THEOREM.

Equal triangles upon the same base and upon the same side of it, are between the same parallels.

Let the equal triangles ABC, DBC, be upon the same base BC, and upon the same side of it; they are between the same parallels.

B

Join AD. AD is parallel to BC. For, if it is not, through the point A draw (I. 31.) AE parallel to BC, and join EC. Then (I. 37.)

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. But the triangle ABC is equal to the triangle BDC (Hyp.); therefore also

2. The triangle BDC is equal to the triangle EBC, the greater to the less, which is impossible. Therefore

3. AE is not parallel to BC. In the same manner, it can be demonstrated that no other line but AD is parallel to BC; therefore

4. AD is parallel to BC. Wherefore, equal triangles upon, &c.

Q.E.D.

PROP. XL.THEOREM.

Equal triangles upon equal bases in the same straight line, and towards the same parts, are between the same parallels.

Let the equal triangles ABC, DEF, be upon equal bases BC, EF, in the same straight line BF, and towards the same parts ; they are between the same parallels.

A

D

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Join AD. AD is parallel to BC. For, if it is not, through A draw (I. 31.) AG parallel to BF, and join GF. Then (I. 38.)

1. The triangle ABC is equal to the triangle GEF, 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 also

2. The triangle DEF is equal to the triangle GEF, the greater 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 parallel to it but AD; therefore

4. AD is parallel to BF. Wherefore, equal triangles, &c.

Q.E.D.

PROP. XLI.-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; the parallelogram ABCD is double of the triangle EBC.

B

Join AC. Then (I. 37.)

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. But (1. 34.)

2. The parallelogram ABCD is double of the triangle ABC, because the diameter AC divides it into two equal parts; wherefore also

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

Q.E.D.

PROP. XLII.-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. Bisect (I. 10.) BC in E, join AE, and at the point E in the straight line EC make (I. 23.) the angle CEF equal to D; and through A draw (I 31.) AFG parallel to EC, and through C draw CG parallel to EF. Therefore (I. 34. Def.) FECG is a parallelogram.

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And because BE is equal to EC, likewise (I. 38.)

1. The triangle ABE is equal to the triangle AEC, since they are upon equal bases BE, EC, and between the same parallels BC, AG; therefore

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

And likewise (I. 41.)

3. The parallelogram FECG is double of the triangle AEC, because they are upon the same base EC, and between the same parallels EC, AG. Therefore (Ax. 6.)

4. The parallelogram FECG is equal to the triangle ABC, and it has one of its angles CEF equal (Constr.) to the given angle D. Wherefore there has been described a parallelogram FECG equal to a given triangle ABC, having one of its angles CĚF equal to the given angle D. Which was to be done.

PROP. XLIII.-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, FG, the parallelograms about AC, that is, through which 4C passes, and BK, KD, the other parallelograms which make up the whole figure ABCD, which are therefore called the complements. The complement BK is equal to the complement KD.

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Because ABCD is a parallelogram, and AC its diameter, (I. 34.)

1. The triangle ABC is equal to the triangle ADC; and because EKHA is a parallelogram, the diameter of wbich is AK,

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

3. The triangle KGC is equal to the triangle KFC. Then because the triangle AEK is equal to the triangle AHK, and the triangle KGC to KFC; (Ax. 2.)

4. The triangle AEK together with the triangle KGC is equal to the triangle AHK together with the triangle KFC. But the whole triangle ABC is equal to the whole ADC; therefore

5. The remaining complement BK is equal to the remaining complement KD. Wherefore, the complements, &c. Q.E.D.

PROP. XLIV.--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.

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 D.

F

E

K

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Make (I. 42.) the parallelogram BEFG equal to the triangle C, and having the angle EBG equal to the angle D, so that BE be in the same straight line with AB; and produce FG to H; and through A draw (I. 31.) AH parallel to BG or EF, and join HB. Then because the straight line HF falls upon the parallels AH, EF, (I. 29.)

1. The angles AHF, HFE, are together equal to two right angles ; wherefore (Ax. 9.)

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

3. HB, FE, shall meet, if produced ; let them meet in K, and through K draw KL parallel to EA or FH, and produce HA, GB, to the points L, M. Then HLKF is a parallelogram, of which the diameter is ĖK, and AG, ME, are the parallelograms about HR, and LB, BF, are the complements; therefore (I. 43.)

4. LB is equal to BF. But BF is equal to the triangle C; wherefore (Ax. 1.)

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

6. The angle ABM is equal to the angle D. Therefore the parallelogram LB is applied to the straight line AB, is equal to the triangle C, and has the angle ABM equal to the angle D. Which was to be done.

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