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VI. Things which are double of the same, are equal to one another.
VII. Things which are halves of the same, are equal to one another.
VIII. Magnitudes which coincide with one another, that is,
which exactly fill the same space, are equal to one another.
Two straight lines cannot enclose a space.
XII. “ If a straight line meet two, straight lines, so as to
66 make the two interior angles on the same side of " it taken together less than two right angles, these “ straight lines being continually produced, shall at “ length meet upon that side on which are the angles “ which are less than two right angles." See the notes on Prop. 29. of Book 1.
B E late.
PROPOSITION I. PROBLEM.
From the centre A, at the distance AB, describe * the circle BCD, and from the centre B, at the distance BA, describe the circle ACE; and from the point C, in which the circles cut one another, draw the straight lines * CA, CB, to the points *1 Post. . A, B; ABC shall be an equilateral triangle.
Because the point A is the centre of the circle BCD, AC is equal* to AB; and because the point B is the * 15 Deficentre of the circle ACE, BC is equal to BA: but it has nition. been proved that CA is equal to AB; therefore CA, CB, are each of them equal to AB: but things which are equal to the same thing are equal * to one another; - 1st Axtherefore CA is equal to CB: wherefore CA, AB, BC iom. are equal to one another: and the triangle ABC is therefore equilateral, and it is described upon the given straight line AB. Which was required to be done.
PROP. II. PROB.
From a given point to draw a straight line equal to a
given straight line. Let A be the given point, and BC the given straight line; it is required to draw from the point A a straight line equal to BC.
From the point A to B draw* the straight line AB; * 1 Post. and upon it describe * the equilateral triangle DAB, * 1. 1. and produce * the straight lines DA, DB, to E and F; * 2 Post. from the centre B, at the distance BC describe * the cir- *3 Post. cle CGH, and from the centre D, at the distance DG describe the circle GKL. AL shall be equal to BC.
Because the point B is the centre of the circle CGH, BC is equal * 10 BG; and because D is the centre of the circle GKI., DL is equal to DG; and + DA, DB, parts of them, are equal; therefore the remainder AL is equal to the remainder * BG: but it has been shewn,
* 15 Def.
that BC is equal to BG; wherefore AL and BC are each of them equal to BG: and things that are equal to the same thing are equalt to one another; therefore the straight line AL is equal to BC. Wherefore from the given point A a straight line AL has been 'drawn equal to the given straight line BC. Which was to be done.
PROP. III. PROB.
From the greater of two given straight lines to cut off a
part equal to the less. Let AB and C be the two given straight lines, whereof AB is the greater. It is required to cut off from AB,
Е В the greater, a part equal to C, the less.
From the point A draw * the straight line AD equal to C; and from the centre A, and at the distance AD describe * the circle DEF: AE shall be equal to c.
Because A is the centre of the circle DEF, AE is equal to AD; but the straight line C is likewise equal to + AD; whence AE and C are each of them equal to AD); wherefore the straight line AE is equal to * C, and from AB, the greater of two straight lines, a part AE has been cut off equal to C the less. Which was to be done.
* 3 Post.
* 15 Def. + Constr.
PROP. IV. THEOR.
If two triangles have two sides of the one equal to two
sides of the other, each to each ; and have likewise the angles contained by those sides equal to one another ; they shall likewise have their bases, or third sides, equal ; and the two triangles shall be equal, and their other angles shall be equal, each to each, viz. those to which the equal sides are opposite.
Let ABC, DEF be two triangles, which have the two sides AB, AC equal to the two sides DE, DF, each to each, viz. AB to DE, and AC to DF; and the angle B BAC equal to the angle EDF: the base BC shall be equal to the base EF; and the triangle ABC to the triangle DEF; and the other angles
to which the equal sides are opposite, shall be equal
For, if the triangle ABC be applied to DEF, so that
with the point E, because AB is equal to DE: and AB coinciding with + Hyp. DE, AC shall coincide with DF, because the
angle BAC is equal t to the angle EDF; wherefore also the + Hyp. point C shall coincide with the point F, because the straight line AC + is equal to DF: but the point B was + Hyp. proved to coincide with the point E: wherefore the base BC shall coincide with the base EF; because, the point B coinciding with E, and C with F, if the base BC does not coincide with the base EF, two straight. lines would inclose a space, which is impossible.* There- * 10 Ax. fore the base BC coincides with the base EF, and therefore is equal t to it. Wherefore the whole triangle ABC + 8 Ax. coincides with the whole triangle DEF, and is equal to it; and the other angles of the one coincide with the remaining angles of the other, and are equal to them, viz. the angle ABC to the angle DEF, and the angle ACB to DFE. Therefore, if two triangles have two sides of the one equal to two sides of the other, each to. each, and have likewise the angles contained by those sides equal to one another, their bases shall likewise be equal, and the triangles shall be equal, and their other angles to which the equal sides are opposite shall be equal, each to each. Which was to be demonstrated.
PROP. V. THEOR.
The angles at the base of an isosceles triangle are equal
to one another; and if the equal sides be produced, the
Let ABC be an isosceles triangle, of which the side
In BD take any point F, and from AE the greater, cut off AG equal* to AF, the less, and join FC, GB.
Because AF is equal to + AG, and AB to | AC, the + Constr. two sides FA, AC are equal to the two GA, AB, each $ Hyp. to each: and they contain the angle FAG common to
** 3. 1.
the two triangles AFC, AGB; therefore
COROLLARY.-Hence every equilateral triangle is also equiangular.
* 4. 1.
PROP. VI. THEOR.
If two angles of a triangle be equal to one another, the
sides also which subtend, or are opposite to, the equal angles, shall be equal to one another.
Let ABC be a triangle having the angle ABC equal to the angle ACB: the side AB shall be equal to the side AC.
For, if AB be not equal to AC, one of them is greater than the other: let AB be the greater; and from it cut * off DB equal to AC, the less, and join DC: therefore, because in the triangles DBC, ACB, DB is equal to AC, and BC common to both, the two sides, DB, BC are equal to the two AC, CB, each to
* 3. 1.