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Book 1. Let the parallelogram ABCD and the triangle EPC be upon

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

Join AC; then the triangle ABC is equal to the triangle EBC, because they are upon the fame base BC, and

between the fame parallels BC, AE. b. 34. s. but the parallelogram ABCD is doubleb of the triangle ABC, because the dia

B. c meter AC divides it into two equal parts; wherefore ABCD is also double of the triangle EBC. therefore if a parallelogram, &c. Q. E. D.

a. 37. 1.

PRO P. XLII. PROB.

To
"O describe a parallelogram that shall be equal to

a given triangle, and have one of its angles cqual to a given rectilineal angle.

a. 10. 1.

b. 23. 1. c. 3*. I.

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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 * BC in E, join AE, and at the point E in the straight line EC make the angle CEF equal to D; and thro' A draw AG parallel to EC, and thro'C draw

AF
CG parallel to EF. therefore
FECG is a parallelogram. and be-

caufe BE is equal to EC, the trid. 38. 1. angle ABE is likewise equal d to the triangle AEC, since they are

D
upon equal bases BE, EC and be-
tween the fame parallels BC,AG;

B E C
therefore the triangle ABC is dou-

ble of the triangle AEC. and the parallelogram FECG is likewise c. 41. 1. double e of the triangle AEC, because it is upon the fame bafe,

word between the fame parallels. therefore the parallelogram FECG is equal to the triangle ABC, and it has one of its angles CEF qual to the given angle D. wherefore there has been described a

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parallelogram FECG equal to a given triangle ABC, having one of Book I. its angles CEF equal to the given angle D. Which was to be done.

PROP. XLIII. THEO R.
The complements of the parallelograms which are

about the diameter of any parallelogram, are equal to one another.

L2t ABCD be a parallelogram, of which the diameter is AC, and EH, FG the parallelograms A H D about AC, that is, thro' which AC passes, and BK, KD thę

IK other parallelograms which

E

E make up the whole figure ABCD, which are therefore called the complements. the complement BK is equal to the complement KD.

B G Because ABCD is a parallelogram, and AC its diameter, the triangle ABC is equal to the triangle ADC, and because EKHA 2. 34. 1 is a parallelogram, the diameter of which is AK, the triangle AEK is equal to the triangle AHK. by 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.

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T O a given straight line to apply a parallelogram,

which shall be equal to a given triangle, and have one of its angles cqual 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 have ing an angle equal to D.

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c. 29. I.

Book I. Make a the

F E

К. parallelogram 2. 42. I.

BEFG equal to
i he' triangle C
and having the

G

M angle EBG

B
qual to the angle
D, so that BE
be in the same

H А. I straight line with AB, and produce FG to H; and thro' A draw b. 31. 1.

b Al parallel to BG or EF, and join HB. then because the straight line HF falls upon the parallels AH, EF,the angles AHF, HFE are together equal 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 d. 12. Ax. fame fide less than two right angles, do meet d if produced far

enough. therefore HB, FE fall meet, if produced; let them meet in K, and thro' 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 e

quale to BF. but BF is equal to the triangle C; wherefore LB is f. 15.1. equal to the triangle C. and because the angle GBE is equals

to the angle ABM, and likewise to the angle D; 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.

e. 43. I.

PRO P. XLV. PRO B.

'O describe a parallelogram equal to a given rectili

neal 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 GH apply b the parallelogram GM equal

2. 42. I.

b. 44. 1.

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C. 29. 1.

to the triangle DBC having the angle GHM equal to the angle E. Book I. 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, A

D F G L KHG are equal to the anges KHG,

E GHM. but FKH, KHG are equal to two right angles; therefore also KHG, GHM are

B 0 K HM 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 sides of it make the adjacent angles equal to two right angles, KH is in the same straight a line with HM. and because the straight line d. 14. 5, HG meets the parallels KM, FG, the alternate angles MHG, HGF are equal“; add to each of these the angle HGL; therefore the angles MHG, HGL are equal to the angles HGF, HGL. but the angles MHG, HGL are equal 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. and because KF is parallel to HG, and HG to ML; KF is parallel e to ML. and KM, e. 30. 1. 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 to the given straight line, a parallelogram equal b. 44. ?, to the first triangle ABD, and having an angle equal to the given angle.

CA

Book 1.

PROP. XLVI. PROB.

O describe a square upon a given straight line.

2. II. I.

b. 3. 1. c. 31. I.

d. 34. 1.

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

From the point A draw a AC at right angles to AB; and make b AD equal to AB, and thro’ the point D draw DE parallel to it, and thro' B draw BE parallel to AD. therefore ADEB is a parallelogram ; whence AB is equal d to DE, and AD to BE. but BA is equal to AD; therefore the four

C straight lines BA, AD, DE, EB are equal to one another, and the parallelogram ADEB is equilateral. likewise all D its angles are right angles; because the straight line AD meeting the parallels AB, DE, the angles BAD, ADE are equal o to two right angles; but BAD is a right angle, therefore also ADE is a right angle. but the opposite angles A B of parallelograms are equal d; therefore each of the opposite angles ABE, BED is a right angle ; wherefore the figure ADEB is rectangular. and it has been demonstrated that it is equilateral ; it is therefore a square, and it is described upon the given straight line AB. Which was to be done.

CoR. Hence every parallelogram that has one right angle has all its angles right angles.

6. 29. 1.

PROP. XLVII. THEOR.
IN any right angled triangle, the square which is de-

scribed upon the side fubtending the right angle, is equal to the squares described upon the sides which contain the right angle.

Let ABC be a right angled triangle having the right angle BAC; the square described upon the side BC, is equal to the squares described upon BA, AC.

On BC describe the square BDEC, and on BA, AC the squares

2. 46. t.

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