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Book I. Because the point B is the center of the circle CGH, BC is
equal to BG. and because D is the center of the circle GKL, DL e. 15. Def.
is equal to DG, and DA, DB parts of them are equal; therefore f.
the remainder AL is equal to the remainder f BG. but it has been shewn that BC is equal to BG; wherefore AL and BC are each of them equal to BG, and things that are equal to the fame are equal 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.
PRO P. III. PROB.
off a part equal to the less.
from the center A, and at the difb. 3. Post.
tance AD describe the circle DEF. and because A is the center of the circle DEF, AE shall be equal to AD. but the straight
line C is likewise equal to AD, whence AE and C are each of C. 1. Ax. 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.
PROP. IV. THEOREM.
two sides of the other, each to each; and have likewise the angles contained by those fides equal to one another: they shall likewise have their bases, or third fides, 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 Gdes DE, DF, each to each, viz. AB to DE,
and AC to DF; and the an
D gle 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, thall be equal, each B CE to each, viz. the angle ABC to the angle DEF, and the angle ACB to 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 DF, because the angle BAC pullon is equal to the angle EDF.' wherefore also the point C shåll coinçide with the point F, because the traight line AC is equal to DF. but the point B coincides 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. Therefore the base BC shall coincide with the a. 10. Ax, base EF, and be equal to it. Wherefore the whole triangle ABC shall coincide with the whole triangle DEF, and be equal to it; and the other angles of the one shall coincide with the remaining angles of the other, and be equal to them, viz. the angle ABC. to the angle DEF, and the angle ACB to DFE. Therefore if two triangles hảve two sides of the one equal to two sides of the other, each to each, and have likewise the angles contained by those fides equal to one another; their bases shall likewise be equal, and the triangles be equal, and their other angles to which the equal fides are opposite, shall be equal, each to each. Which was to be demonstrated.
PRO P. V. THEOR.
equal to one another; and if the equal fides be produced, the angles upon the other side of the base shall be equal.
Let ABC be an Isosceles triangle, of which the Ạde AB is equal
Book I. to AC, and let the straight lines AB, AC be produced to D and m 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 a. 3. 1. equal to AF, the less, and join FC, GB.
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 tri
FC is equal b to the base GB, and the
E. whole AG, of which the parts AB, AC c. 3. Ax. are equal; the remainder BF shall be 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 is equal to the angle CGB; and the base BC is common to the two triangles BFC, CGB; wherefore the triangles are equal , and their remaining angles, each to each, to which the equal fides are opposite. therefore the angle FBCis 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.
F two angles of a triangle be equal to one another,
the fides 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 Book I. angle ACB; the side AB is also 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, a. 3. I. the less, and join DC. therefore because in
А. the triangles DBC, ACB, DB is equal to AC, and BC common to both, the two
D fides DB, BC are equal to the two AC, CB, each to each; and the angle DBC is equal to the angle ACB; therefore the base DC is equal to the base AB, and the triangle DBC is equal to the triangle b ACB, the less to the greater; which B
b. 4. I. is absurd. 'Fherefore AB is not unequal to AC, that is, it is equal Wherefore if two angles, &c. Q. E. D.
Cok. Hence every equiangular triangle is also equilateral.
PRO P. VII. THEOR.
there cannot be two triangles that have their fides which are terminated in one extremity of the bafe equal to one another, and likewise those which are terminated in the other extremity.
If it be possible, let there be two triangles ACB, ADB upon the same base AB, and upon the same side of it, which have their sides CA, DA, terminated in the extremity A of the base, equal to one another, and likewise their fides CB, DB that are terminated in B.
Join CD; then, in the case in which the Vertex of each of the triangles is without the other triangle, because AC is equal to AD, the angle ACD is equal a to the angle ADC. but the
a. 5. I. angle ACD is greater than the angle BCD, therefore the angle ADC is greater also than BCD; much more then is A
3 the angle BDC greater than the angle BCD. again, because ÇB is equal to DB, the angle BDC is equal a to the angle BCD; but it has been demonstrated to be greater than it; which is impossible.
2. 5. I.
But if one of the Vertices, as D, be within the other triangle
Therefore upon the fame base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated in the other extremity. Q. E. D.
two triangles have two sides of the one equal to
two sides of the other, each to each, and have like. wise their bases equal; the angle which is contained by the two sides of the one shall be equal to the angle contained by the two fides equal to them, of the other.
Let ABC, DEF be two triangles having the two sides AB, AC
For if the trian-