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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 AB a parallelogram equal to the triangle C, and having an angle equal to D. Make (42. 1.) the parallelogram BEFG equal to the triangle
L C, having the angle EBG equal to the angle D, and the side BE in the same straight line with AB : produce FG to H, and through A draw (31. 1.) AH 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 (29. 1.) 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 if produced (Cor. 29. 1.): Therefore HB, FE will meet, if produced ; let them meet in K, and
l througb 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 (43. 1.) to BF : but BF is equal to the triangle C; wherefore LB is equal to the triangle C; and because the angle GBE is equal (15. 1.) to the angle ABM,
; and likewise to the angle D ; the angle ABM is equal to the angle D: Therefore the parallelogram LB, which 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 hav.
ing 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 (42. 1.) the parallelogram FH equal to the triangle ADB, and having the angle HKF equal to the angle E; and to the straight line GH (44. 1.) apply the parallelogram GM equal to the triangle DBC, having the angle GHIM 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 anglı KHG; herefore the angles FKH, KHG are equal to the angles KHG, GHM ; but FKH, KHG are equal (29 1.) to two right angles ; therefore also KHG, GHM are equal to two right angles : and because at the point D
B K Н. М. # in the straight line GH, the two straight lines KH, HM, upon the opposite sides of GH, make the adjacent angles equal to two right angles, KH, is in the same straight line (14. 1.) with HM. And because the straight line HG meets the parallels KM, FG, the alternate angles MHG, HGF are equal (29. 1.); add to each of these the angle HGL ; therefore the angles MHG. HGL, are equal o the angles HGF, HGL: But the angles MHG, HGL, are equal (29. 1.) 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 (30. 1.) to ML; but 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 rectili. neal angle, and shall be equal to a given rectilineal figure, viz. by applying (44. 1.) 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 AB.
From the point A draw (11. 1.) AC at right angles to AB ; and make (3. 1.) AD equal to AB, and through the point D draw DE pa
rallel (31. 1.) to AB, and through B draw BE parallel to AD; therefore ADEB is a parallelogram : Whence AB is equal (34. 1.) to DE and AD to BE: but BA is equal to AD; C therefore the four straight lines BA, AD, DE, EB are equal to one another, and the parallelogram ADEB is equilateral ;
E it is likewise rectangular ; for the straight line AD meeting the parallels, AB, DE, makes the angles BAD, ADE equal (29. 1.) to two right angles ; but BAD is a right angle ; therefore also ADE is a right angle now the opposite angles of parallelograms are equal (34. 1.); therefore each of the opposite angles ABE, BED is a right an- А
B gle; 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.
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
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 (46. 1.) the square BDEC, and on BA, AC the squares GB, HC ; and through A draw (31. 1.) AL parallel to BD or CE, and join AD, FC; then, because each of the angles BAC, BAG is a right angle (25. def.), the two straight lines AC, AG upon
G the opposite sides of AB, make with it at the point A the adja
н cent angles equal to two right an- E gles ; therefore CA is in the same straight line (14. 1.) with AG ;
K for the same reason, AB and AH are in the same straight line. Now because the angle DBC is
B equal to the angle FBA, each of them being a right angle, adding to each the angle ABC, the whole angle DBA will be equal (2. Ax.) to the whole FBC; and because the two sides AB, BD, are equal to the two FB, BC each, to each
LE and the angle DBA equal to the
angle FBC, therefore the base AD is equal (4. 1.) to the base FC, and the triangle ABD to the triangle FBC. But the parallelogram BL is double (41. 1.) of the triangle ABD, because they are upon the same base, BD, and between the same parallels, BD, AL; and the square GB is double of the triangle BFC, because these also are upon the same base FB, and between the same parallels FB, GC. Now the doubles of equals are equal (6. Ax.) to one another ; 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
the other two sides of it; the angle contained by these two sides is a right angle. If the square described upon BC, one of the sides of the triangle ABC, be equal to the squares upon the other sides BA, AC, the angle BAC is a right angle.
From the point A draw (11. 1.) AD at right angles to AC, and make AD equal to BA, and join DC. Then because DA is equal to
, AB, the square of DA is equal to the square of AB: To each of these add the
of AC; therefore the squares of DA, AC are equal to the squares of BA, AC. But the square of DC is equal (47. 1.) to the squares A Α. of DA, AC, because DAC is a right angle ; and the square of BC, by hypothesis, is equal to the squares of BA, AC; therefore, the square of DC is equal to the square of BC ; and therefore also the side DC is equal to the
С side BC. And because the side DA is equal to AB, and AC common to the two triangles DAC, BAC, and the base DC likewise equal to the base BC, the angle DAC is equal (8. 1.) to the angle BAC : But DAC is a right angle; therefore also BAC is a right angle. Therefore, if the square, &c. Q. E. D.
1. VERY right angled parallelogram, or rectangle, is said to be con
tained by any two of the straight lines which are about one of the right angles
“ Thus the right angled parallelogram AC is called the rectangle es contained by AD and DC, or by AD and AB, &c. For the sake of brevity, instead of the rectangle contained by AD and DC, we shall " simply say the rectangle AD.DC, placing a point between the two “ sides of the rectangle Also, instead of the square of a line, for “instance of AD, we shall frequently in what filliws write AD2.”
" The sign + placed between the names of two magnitudes, signi"fies that those magnitudes are to be added together, and the sign "-placed between them, signifies that the latter is to be taken away « from the former.'
“ The sign = signifies, that the things between which it is placed " are equal to one another.”
II. In every parallelogram, any of
Α. E the parallelograms about a ameter, together with the two complements, is called a Gno
“ Thus the parallelogram HG, together with the "complements AF, FC, is the TI
K “gnomon of the parallelogram
AC, This gnomon may also, “ for the sake of brevity, be
B G “called the gnomon AGK or «EHC."