Book I.

PROP. XL. THEOR.

in fame

straight line, and towards the fame parts,

between the fame parallels.

Let the equal triangles ABC, DEF be upon equal bases

BC, EF, in the fame

A

D

G

they are between the

fame parallels.

a 31. 1.

therefore parallel to BF.
Q. E. D.

Join AD; AD is parallel to BC: For, if it B is not, through A draw a AG parallel to BF, and join GF. The triangle ABC is b 38. 1. equal b to the the triangle GEF, because they are upon equal bafes BC, EF, and between the fame parallels BF, AG: But the triangle ABC is equal to the triangle DEF; therefore also the triangle DEF is equal to the triangle GEF, the greater to the less, which is impossible: Therefore 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; AD is Wherefore equal triangles, &c.

C

F

PROP. XLI. THEOR.

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

Let the parallelogram ABCD and the triangle EBC be upon the fame base BC, and between the fame parallels BC,

diameter AC divides it into two equal parts; wherefore ABCD is alfo double of the triangle EBC. Therefore, if a

parallelogram, &c. Q. E. D.

Toda

PROP. XLII. PROB.

O describe a parallelogram that shall be equal 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 a BC in E, join AE, and at the point E in the a 10. 1. straight line EC make b the angle CEF equal to D; and b 23. 1. through A draw c AG parallel to EC, and through C draw c 31. 1.

CGc parallel to EF: There

fore FECG is a parallelo- A

gram: And because BE is

equal to EC, the triangle

ABE is likewise equal d. to

d 38. 1.

the triangle AEC, fince

they are upon equal bases BE, EC, and between the

D

fame parallels BC, AG;

therefore the triangle ABC

is double of the triangle AEC. And the parallelogram FECG

is likewise double e of the triangle AEC, because it is up- e 41. 1.

on the fame base, and between the fame parallels: Therefore

the parallelogram FECG is equal to the triangle ABC, and

it has one of its angles CEF equal to the given angle D:

Wherefore

Book I. Wherefore there has been described a parallelogram FECG equal to a given triangle ABC, having one of it angles CEF equal to the given angle D. Which was to be done.

T

PROP. XLIII. THEOR.

HE 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; let EH, FG the parallelograms about AC, that is, thro'

which AC passes, and BK, KD

the other parallelograms which

make up the whole figure
ABCD, which are therefore E

called the complements: The
complement BK is equal to the
complement KD.

Because ABCD is a parallelogram, and AC its diameter, B

2 34. 1. the triangle ABC is equal a to

the triangle ADC: And, because EKHA is a parallelogram, the diameter of which is AK, the triangle AEK is equal to the triangle AHK: For the fame reason, 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; 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 the remaining complement BK is equal to the remaining complement KD. Wherefore the complements, &c. Q. E. D.

T

PROP. XLIV. PROB.

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.

Book I.

Make a the parallelogram BEFG equal to the triangle C, & 42. 1.

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 by AH parallel to BG or EF, and join b 31. 1. HB. Then because the straight line HF falls upon the parallels AH, EF, the angles AHF, HFE, are together equal c c 29. I. 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 fide less than two right angles, do meet if produced far enough: Therefore 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 HK, and AG, ME are the parallelograms about HK; and LB, BF are the complements; therefore LB is equal d to BF: But BF is e- d 43. Y. qual to the triangle C; wherefore LB is equal to the triangle C; and because the angle GBE is equale to the angle e 15. 1, ABM, and likewise to the angle D; the angle ABM is equal to the angle D: Therefore the parallelogram

LB

Book I. LB is applied to the straight line AB, is equal to the triangle C, and has the angle ABM equa, to the angle D: Which was to be done.

a 42. I.

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 a the parallelogram FH equal to the triangle ADB, and having the angle HKF equal to the angle E; and to the straight line GHb apply 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 FKH, GHM, the angle FKH is equal to GHM; add to each of these the angle KHG; therefore the angles FKH, KHG are equal to the angles KHG, GHM;

€ 29. 1.

but FKH, KHG are equal c to two right angles; there

14.1.

fore also KHG, GHM are equal to two right angles: and because at the point H in the straight line GH, the two straight lines KH, HM, upon the opposite fides of GH, make the adjacent angles equal to two right angles, KH is in the same straight line d with HM. And because the straight line HG meets the parallels KM, FG, the alternate angles MHG, HGF are equal c; add to each of these the angle HGL; therefore

the angles MHG, HGL, are equal to the angles HGF, HGL :

But