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PROPOSITION I. PROBLEM.
From the centre A, at the dis* 3 Pos- tance AB, describe * the circle
BCD, and from the centre B, at
C, in which the circles cut one
points A, B; ABC shall be an equilateral triangle.
Because the point A is the centre of the circle BCD, * 15 De- AC is equal* to AB; and because the point B is the finition.
centre of the circle ACE, BC is equal to BA: but it
PROP. II. PROB.
given straight line.
From the point A to B draw* the straight line AB;
and upon it describe* the equilateral triangle DAB, * 2 Post. and produce* the straight line DB to E; from the * 3 Post. centre B, at the distance BC, describe * the circle
CFH, and from the centre D, at the distance DF,
* Ist Axiom.
* 1 Post.
describe the circle FGK: produce the straight line
* 15 Def. of the circle CFH, BC is equal* to BF; and because D is the centre of the circle FGK, DL is equal to DF, and DA, DB, parts of them, are equal; therefore the remainder AL is equal * to the
* 3 Ax. remainder BF: but it has been shown, that BC is equal to BF; wherefore AL and BC are each of them equal to BF; and things that are equal to the same thing are equal * to one another; * 1 Ax. 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.
a part equal to the less.
c cut off from AB, 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: * 3 Post. then AE shall be equal to C. Because A is the centre of the circle DEF, AE is equal* to AD; but the * 15 Def. 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 therefore from * 1 Ax. AB, the greater of two straight lines, a part AE has been cut off equal to C the less. Which was to be done.
* 2. I.
PROP. IV. THEOREM.
sides of the other, each to each; and have likewise
Let ABC, DEF be two tri- A angles, 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 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, each to each, viz. the angle ABC to the angle DEF, and the angle ACB to the angle DFE.
For, if the triangle ABC be applied to DEF, so that the point A may be on D, and the straight line AB upon DE; the point B shall coincide with the point E, because AB is equal to DE; and AB coinciding with DE, AC shall coincide with DF, because the angle BAC is equal to the angle EDF: wherefore, also, the point C shall coincide with the point F, because the straight line AC is equal to DF: but the point B was 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 did not coincide with the base EF, two straight * 10 ix. lines would enclose a space, which is impossible.*
Therefore the base BC coincides with the base EF, and is therefore equal to it. Wherefore the whole triangle ABC 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 angles upon the other side of the base shall be equal.
Let ABC be an isosceles triangle, of which the side AB is equal to AC, and let the straight lines AB, AC, be produced to D and E: the angle ABC shall be equal to the angle ACB, and the angle CBD to the angle BCE.
In BD take any point F, and from AE the greater, cut off AG equal * to AF, the less, and join FC, GB.
* 3. 1. Because AF is equal to AG, and AB to AC, the two sides FA, AC, are equal to the two GA, AB, each to each; and they contain the angle FAG common to the two triangles AFC, AGB; there
* 4. 1. fore the base FC is equal * to the base GB, and the ngle AFC to the triangle AGB; and the remain
* 4. 1. ing angles of the one are equal * to the remaining angles of the other, each to each, to which the equal sides are opposite; viz. the angle
* 4. 1.
ACF to the angle ABG, and the angle AFC to the angle AGB: and because the whole AF is equal to the
whole AG, of which the parts AB, AC, are equal; the * 3 Ax. remainder BF is equal * to the remainder CG; and
FC was proved to be equal to GB; therefore the two sides BF, FC are equal to the two CG, GB, each to each; and the angle BFC has been proved to be equal to the angle CGB; wherefore the two triangles BFC, CGB are equal,* and their remaining angles are equal, each to each, to which the equal sides are opposite; therefore the angle FBC is equal to the angle GCB, and the angle BCF to the angle CBG. And, since it has been demonstrated, that the whole angle ABG is equal to the whole ACF, the parts of which, the angles CBG, BCF are also equal; the remaining angle ABC is therefore equal to the remaining angle ACB, which are the angles at the base of the triangle ABC: and it has also been proved that the angle FBC is equal to the angle GCB, which are the angles upon the other side of the base. Therefore the angles at the base, &c. Q. E. D.
COROLLARY. Hence every equilateral triangle is also equiangular,
PROP. VI. THEOR.
sides also which subtend, or are opposite to, the equal
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
* 3. 1.