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* 12 Ax.
EUCLID'S ELEMENTS. st 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 same side 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 parallelograms about HK; and LB, BF are the complements: therefore LB is equal * 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 * to the angle ABM, and likewiset to the angle D; the angle ABM is equal to the angle D. Therefore, to the straight line AB the parallelogram LB is applied, equal to the triangle C, and having the angle ABM equal to the angle D. Which was to be done.
* 43. 1.
+ Constr, +1 Ax.
+ Constr. 11 Ax.
PROP. XLV. PROB.
* 42. 1.
* 44. 1.
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 * the parallelogram FH equal to the triangle ADB, and having the angle FKH equal to the angle E; and to the straight line GH apply* the parallelogram GM equal to the triangle DBC, having the angle GHM equal to the angle E: the figure FKML shall be the parallelogram required.
Because the angle E is equal + to each of the angles FKH, GHM, the angle FKH is equal t to GHM: add to each of these the angle KHG; therefore the angles FKH, KHG, are equal + to the angles KHG, GHM: but FKH, KHG are equal * to two LE right angles; therefore alsoKHG, GHM, are equalt to two right angles: and because at the point
CK HM H in the straight line GH, the
of Constr. + 1 Ax.
+ 2 Ax,
* 29. 1.
+ 1 Ax.
* 30. 1.
two straight lines KH, HM upon the opposite sides
† with GL: and † 14. 1.
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 to the first triangle ABD, and having an angle equal to the given angle.
+ Def. 34.
* 44. 1.
PROP. XLVI. PROB.
* 31. 1.
To describe a square upon a given straight line.
From the point A draw* AC at right angles to AB; - 11.1.
+ Constr. straight lines BA, AD, DE, EB, are equal t. to one another, and the parallelogram ADEB is equilateral : likewise all its angles are right angles; for, since the straight line AD meets the parallels AB, DE, the angles BAD, ADE are
+ 1 Ax.
* 29. 1.
+ Constr. +
34. 1. + 1 Ax.
equal* to two right angles: but BAD is at right angle; therefore also ADE is at right angle: but the opposite angles of parallelograms* are equal; therefore each of the opposite angles ABE, BED is at right angle ; wherefore the figure ADEB is rectangular : and it has
been demonstrated that it is equilateral ; it is therefore + 50 Def. at 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.
# 46. 1.
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 shall be equal to the squares described upon BA, AC.
On BC describe * the square BDEC, and on BA, AC the squares GB, HC; and through A draw * AL
parallel to BD, or CE, and join AD, FC. Then, + Нур. because the angle BAC is a f right angle, and that the * 30 Def. angle BAG is also a * right angle,
the two straight lines AC, AG upon
line. And because the angle DBC + 11 Ax.
is equalt to the angle FBA, each of + 30 Def. them being a right + angle, add to each the angle ABC;
therefore the whole angle DBA is equal * to the whole † 30 Def.
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 * to the base FC, and the triangle ABD to the triangle FBC: now the parallelogram BL is double * 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
* 14. 1.
* 41. 1.
between the same parallels FB, GC: but the doubles of equals are equal * to one another; therefore the * 6 Ax. parallelogram BL is equal to the square GB. In the same manner, by joining AE, BK, it can be demonstrated, that the parallelogram CL is equal to the square HC: therefore the
whole square BDEC is equal+ + 2 Ax. to the two squares GB, HC: and the square BDÈC is described upon the straight line BC, and the squares GB, HC upon BA, AC; therefore 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.
Let 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 shall be a right angle. From the point A draw*
AD at right angles to AC, * 1.1. and maket AD equal to BA, and join DC. Then, + 3. 1. because DA is equal to AB, the square of DA is equal to the square of AB: to each of these add the square of AC; therefore the squares of DA, AC are equalt to + 2 Ax. the squares of BA, AC: but the square of DC is equal * to the squares of DA, AC, because DAC is at right angle; and the
+ Constr. square of BC, by hypothesis, is equal to the squares of BA, AC; therefore the square of DC is equalt to the square of
☆ 1 Ax. 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 + Constr. DAC, BAC, the two DA, AC are equal to the two BA, AC each to each; and the base DC has been proved equal to the base BC; therefore the angle DAC is equal* to the angle BAC: but DAC is at right angle; * 8. 1. therefore also BAC is at right angle. Therefore, if Constr. the square, &c.
† 1 Ax.
Q. E. D.
Every right-angled parallelogram, or rectangle, is said to be contained
by any two of the straight lines which contain one of the right angles.
In every parallelogram, any of the parallelograms about
a diameter, together with the two
K к gnomon,
BG briefly expressed by the letters • AGK, or EHC, which are at the opposite angles of the parallelograms which make the gnomon.
PROP. I. THEOR.
If there be two straight lines, one of rohich is divided into
any number of parts; the rectangle contained by the two straight lines, is equal to the rectangles contained by the undivided line, and the several parts of the divided line.
Let A and BC be two straight lines; and let BC be divided into any parts in the points D, E: the rectangle