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Make (42. 1.)
K the parallelogram BEFG equal to the triangle C, and hav.
D ing the angle EBG
G equal to the angle
B D, so that BE be in с the same straight line with AB, and
L 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 (12. Ax.) 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 LM: 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 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.
Join DB, and describe (42. 1.) the parallelogram FH equal to
. See Note,
the triangle ADB, and having the angle HKF equal to the angle E; and to the straight line GH apply (44. 1.) the parallelogram GM equal to the triangle DBC, having the angle GHM equal to the angle E; and because the angle É 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 an- A
F G L gles KHG, GHM; but FKH, KHG are equal
E (29. 1.) to two right angles: therefore also KÉG, GHM equal to two right angles; and because at the point H in the
С K 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 (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 to the angles HGF, HGL ; but the angles HGM, HGL are equal (29. 1.) to two right -angles ; wheresore 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; 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 réctilineal 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 parallel (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; C but BA is equal to AD; therefore, the four straight lines BA, AD, DE, EB are equal to one another, and the parallelogram ADEB is equilateral, likewise all its angles are right angles : because the
E straight lines AĎ meeting the parallels AB, DE, the angles BAD, ADE are equal (29. 1.) to two right angles : but BAD is a right angle; therefore also ADE is a right angle ; but the opposite angles of parallelograms are equal, (34. A
B 1.) 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.
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 (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,
G (30. def.) the two straight lines AC, AG upon the opposite sides of AB, make with it at
F the point A the adjacent angles equal to two right angles :
K therefore CA is in the same straight line (14. 1.) with AG: for the same 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,
D L E and the whole angle DBA is equal (2. Ax.) to the whole FBC : and 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 (4. 1.) to the base FC, and the triangle ABD to the triangle FBC: now 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 FBC, because these also are upon the same 'base FB, and between the same parallels FB, GC. But 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 mander, 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.
If the square described upon BC, one of the sides of the triangle ABC, be equal to the squares upon the other sides BA, AČ, 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
D the square of AB: to each of these add the square 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 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 B
с of BC; and therefore also the side DC is equal to the side BC. And because the side DA is equal to AB, and AC common to the two triangles DAC, BAC, the two DA, AC are equal to the two BA, AC: and the base DC is equal to the base BC: therefore the angle DAC is equal (s. 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.