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be upon the same base BC, and

DE between the same parallels BC, AE ; the parallelogram ABCD is double of the triangle EBC.

Join Ac; then the triangle ABC is equal (1. 37.) to the triangle EBC, because they are upon the same base BC, and between the same parallels BC, A E. But the parallelogram ABCD is double (1. 34.) of the triangle ABC, because the diameter AC divides it into two equal parts; wherefore ABCD is also double of the triangle EBC. Therefore if a parallelogram, &c. Q. E. D.

PROP. XLII. PROB. 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 A B C be the given triangle, and the given rectilineal angle. It is required to describe a parallelogram that shall be equal to the given triangle A B C, and have one of its angles equal to D.

BISECT (1. 10.) Bc in E, join a E, and at the point E in the straight line Ec make (1. 23.) the angle CEF equal to D; and through a draw (1. 31.) AG parallel to E C, and through c draw CG (1. 31.) parallel to E F: Therefore FECG is a parallelogram : And because BE is equal to EC, the triangle A BE is likewise equal (1.38.) to the triangle AEC, since they are upon equal bases BE, EC, and between the same parallels, BC, AG; therefore the triangle A B C is double the triangle A E C. And the parallelogram FECG is likewise double (1. 41.) of the triangle A E C, because it is upon the same base, and

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between the same parallels: Therefore the same parallelogram F ECG is equal to the triangle A B C, and it has one of its angles C E F equal to the given angle D: Wherefore there has been described a parallelogram FECG equal to a given triangle ABC, having one of its angles CEF equal to the given angle D. Which was to be done.

PROP. XLII). TAEOR. The complements of the parallelogram 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 E H, FG the parallelograms about Ac, that

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7 is, through which ac passes, and BK, KD the other parallelograms which make up the whole figure ABCD, which are B therefore called the complements : The complement BK is equal to the complement KD.

Because ABCD is a parallelogram, and Ac its diameter, the triangle A B C is equal (1. 34.) to the triangle ADC: And, because EKHA is a parallelogram, the diameter of which is A K, the triangle A E K is equal to the triangle A HK: By the same reason, the triangle kgc is equal to the triangle KFC: Then, because the triangle A EK is equal to the triangle A HK, and the triangle KGC to KFC; the triangle A E K together with the triangle KGC is equal to the triangle A H K together with the triangle KFC: But the whole triangle A B C is equal to the whole ADC; therefore the remaining complement BK is equal to the remaining complement KD. Wherefore the complements, &c. Q. E. D.

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PROP. XLIV. PROB. 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 A B a parallelogram equal to the triangle c, and having an angle equal to d.

Make (1. 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 (1. 31.) A H parallel to BG or E F, and join 1 B. Then because the straight line HF falls upon the parallels AH, EF, the angles AHF, HFE, are together equal (I. 29.) to two right angles; wherefore the angles BHF, HFE are less than two right angles : But straight lines which with another straight line make the interior angles upon the same side less than two right angles, do meet (Ax. 12.) if produced far enough : Therefore 1 B, FE shall meet, if produced ; let them meet in K, and through a 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 hk, and AG, ME are the parallelograms about hk; and LB, BF are the complements; therefore LB is equal (i. 43.) to BF: But BF is equal to the triangle c; wherefore LB is equal to the triangle

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c; and because the angle GBE is equal (1. 15.) to the angle A BM, and likewise to the angle D (Constr.); the angle ABM is equal to the angle d: Therefore the parallelogram LB is applied to the straight line A B, is equal to the triangle c, and has the angle A BM equal to the angle D: Which was to be done.

PROP. XLV. PROB. 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 rectilineal angle. It is required to describe a parallelogram equal to ABCD, and having an angle equal to E.

Join DB, and describe (1. 42.) the parallelogram FH equal to the triangle ADB, and having the angle HKF equal to the angle E; and to the straight line GH apply (1. 44.) the parallelogram Gm equal to the triangle DBC, having the angle ghm equal to the angle E: and because the angle E is equal to each of the angles FK H, GHM, the angle Fkh is equal to Gam; add to each of these the angle khG; therefore the angles FKH, KHG are equal to the angles KHG, GHM; but FKH, KHG are equal (1. 29.) to two right angles: therefore also KHG, GHM are equal to two right angles ;

and because at the point A u in the straight line GH, the two straight lines KH, Hm upon the opposite sides of it make the adjacent angles equal to two right angles, ku is in the same

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straight line (1. 14.) with hm; and because the straight line ng meets the parallels KM, FG, the alternate angles MHG, HGF are equal (1. 29.): Add to each of these the angle HGL: Therefore the angles MHG, HGL are equal to the angles AGF, HGL: But the angles MHG, HGL are equal (1. 29.) to two right angles; wherefore also the angles HGF, HGL are equal to two right angles, and FG is therefore in the same straight line with GL (1. 14.): And because KF is parallel to ng (Constr.), and ng to ML; KF is parallel (I. 30.) to ML: and KM, FL are parallels; wherefore KFLM is a parallelogram; and because the triangle ABD is equal to the parallelogram HF, and the triangle DBC to the parallelogram GM; the whole rectilineal figure ABCD is equal to the whole parallelogram KFLM; therefore the parallelogram KFLM has been described equal to the given rectilineal figure ABCD, having the angle FKM equal to the given angle E. Which was to be done.

Cor. From this it is manifest how to a given straight line to apply a parallelogram, which shall have an angle equal to a given rectilineal angle, and shall be equal to a given rectilineal figure, viz. by applying (1. 44.) to the given straight line a parallelogram equal to the first triangle ABD, and having an angle equal to the given angle.

PROP. XLVI. PROB. To describe a square upon a given straight line.

Let AB be the given straight line; it is required to describe a square upon a B.

From the point a draw (1. 11.) AC at right angles to AB; and make (1. 3.) AD equal to A B, and through the point o draw de parallel (1. 31.) to AB, and through B draw BE parallel to AD; therefore ADEB

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