« ForrigeFortsett »
FECG equal to a given triangle ABC, having one of its angles Book 1. CEF equal to the given angle D. Which was to be done.
PROP. XLIII. THEOR.
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 A H
D about AC, that is, through which AC passes, and BK, KD the other parallelograms which
E make up the whole figure ABCD, which are therefore called the complements; the complement BK is equal to the complement KD. Because ABCD is a paral- B G
C lelogram, and AC its diameter, the triangle ABC is equal a to the triangle ADC : and a 34. 1. because EKHA is a parallelogram, the diameter of which is AK, the triangle AEK is equal to the triangle AHK : by the same 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.
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 AB a parallelogram equal to the triangle C, and having an angle equal to D.
Book I. Make • the
M equal to the
L straight line b 31. 1. with AB, and produce FG to H; and through A draw 6 AH pa
rallel to BG or EF, and join HB. Then, because the straight
line HF falls upon the parallels AH, EF, the angles AHF, HFE c 29. 1. 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 d 12. Ax. the same side less than two right angles do meet d, 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 parallelograin, of which the diameter is HK, and AG, ME are the
parallelograms about HK; and LB, BF are the complements; e 43. 1. therefore LB is equal e to BF; but BF is equal to the triangle
C; wherefore LB is equal to the triangle C: and because the $ 15. 1. angle GBE is equal f 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.
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. 2 42.1.
Join DB, and describe a the parallelogram FH equal to the
triangle ADB, and having the angle HKF equal to the angle E; b 44.1and to the straight line GH apply b the parallelogram GM equal
to the triangle DBC, having the angle GHM equal to the angle Book I. 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 ; but FKH, A
F G L KHG equal to two right an
c 29. 1.' gles ; therefore
E also KHG, GHM are equal to two right angles; and because at the point H in the
CK H M 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 line d with HM; and because the straight line d 14. 1. HG meets the parallels KM, FG, the alternate angles MHG, HGP are equal c: 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 e to two right an. gles; 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 parallele to ML: and KM, FL are parallels ; wherefore KFLM e 30. 1. 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 b to the given straight line a parallelogram b 44. 1. 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.
a 11. 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 10 AB ; and b 3. 1.
make b'AD equal to AB, and through the point D draw DE o 31. 1. parallel c to AB, and through B draw BE parallel to AD; thered 34. 1. fore ADEB is a parallelogram : whence AB is equal a to DE,
and AD to BE: but BA is equal to с
E line AD meeting the parallels AB,
DE, the angles BAD, ADE are equal e 29. 1. e to two right angles; but BAD is a
right angle; therefore also ADE is a
Cor. Hence every parallelogram that has one right angle has all its angles right angles.
PROP. XLVII. THEOR.
IN any right angled triangle, the square which is described upon the side subtending 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 a the square BDEC, and on BA, AC the
a 46. 1.
squares GB, HC; and through A draw b AL parallel to BD or Book I. CE, and join AD, FC; then, because each of the angles BAC, BAG is a right angles, the
b 31. 1. two straight lines AC, AG,
H upon the opposite sides of AB, make with it at the point A the adjacent angles equal F to two right angles; therefore
K CA is in the same straight line d with AG ; for the same
d 14. 1, reason, AB and AH are in the same straight line ; and because the angle DBC is equal to the angle FBA, each of them being a right angle, add to each the angle ABC, and the whole angle DBA is
D L E equal e to the whole FBC ; and
e 2. Ax because the two sides AB, BD are equal to the two FB, BC, each to each, and the angle DBA equal to the angle FBC; therefore the base AD is equal f to the base FC, and the trian- f 4. 1. ‘gle ABD to the triangle FBC: now the parallelogram BL is double 8 of the triangle ABD, because they are upon the såme g 41. 1. base BD, and between the same parallels, BD, AL; and the square GB is double of the triangle FBC, because these also are upon the same base FB, and between the same parallels FB, GC. But the doubles of equals. are equal h to one another: h 6. Ax. therefore the parallelogram BL is equal to the square GB: and in the same manner, by joining AE, BK, it is demonstrated that the parallelogram CL is equal to the square HC: therefore the whole square BDEC is equal to the two squares GB, HC ; and the square BDEC is described upon the straight line BC, and the squares GB, HC upon BA, AC: wherefore the square upon the side BC is equal to the squares upon the sides BA, AC. Therefore, in any right angled triangle, &c. Q. E. D.
PROP. XLVIII. THEOR.
IF the square described upon one of the sides of a triangle be equal to the squares described upon the ' other two sides of it; the angle contained by these two sides is a right angle.