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Book I. AB to DE, and AC to
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 coincides with the point E; wherefore the base BC shall coincide with the base EF, because the point B coi ciding with E, and C with F, if the base BC does not coincide with the
base EF, two straight lines would enclose a space, which is a 10. Ax. impossible a. Therefore the base BC shall coincide with the
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 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 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 line AB, AC be produced Book I. 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 A E the greater, cut off AG equal a to AF the less, and join FC, GB.
a 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,
A AGB; therefore the base FC is equal to the base GB, and the tri
b 4. 1. angle AFC to the triangle AGB; and the remaining angles of the one are equal to the remaining angles of the other, each to each, to which
C the equal sides are opposite; viz. the angle ACF to the angle ABG, F
G and the angle AFC to the angle AGB: And because the whole AF is equal to the whole AG, of which D
E the parts AB, AC, are equal; the remainder BF shall be equal to the remainder CG; and FC c. 3. Ax. 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 these triangles are equal b, and
their remaining angles, 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.
If two angles of a iriangle 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 is also equal to the side AC.
For, if AB be not equal to AC, one of them is greater than a 3. ). the other: Let AB be the greater, and from it cut a off DB equal to AC, the less, and join DC;
DC is equal to the base AB, and the
C is absurd. Therefore AB is not unequal to AC, that is, it is equal to it. Wherefore, “ if two angles," &c. Q. E. D.
Cor. Hence every equiangular triangle is also equilateral.
PROP. VII. THEOR.
Upon the same base, and on the same side of it,
If it be possible, let there be two triangles ACB, ADB, up-
a 5. 1.
But if one of the vertices, as D, be within the other triangle Book I. ACB; produce AC, AD to E, F; therefore, because AC is equal to AD in the
E triangle ACD, the angles ECD, FDC
F upon the other side of the base CD are equal a to one another, but the
a 5. 1. angle ECD is greater than the angle BČD; wherefore the angle FDC is likewise greater than BCD; much more then is the angle BDC greater than the angle BCD. Again, because CB is equal to DB, the angle BDCA
B is equal a to the angle BCD; but BDC has been proved to be greater than the same BCD; which is impossible. The case in which the vertex of one triangle is upon a side of the other, needs no demonstration. Therefore 66
upon the same 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.
PROP. VIII. THEOR.
If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise their bases equal, the angle which is contained by the two sides of the one shall be equal to the angle contain: ed by the two sides equal to them of the other.
Let ABC, DEF be two triangles, having the two sides AB,
For, if the triangle ABC be applied to DEF, so B CE
F that the point B be on E, and the straight line BC upon EF; the point C shall
Book I. also coincide with the point F; because BC is equal to EF;
therefore BC coinciding with EF, BA and AC shall coincide
sides terminated in the other extremity: But this is imposa 7. 1.
sible a ; therefore, if the base BC coincides with the base EF,
wherefore likewise the angle BAC coincides with the angle b 8. Ax. EDF, and is equalo to it. Therefore “ if two triangles,” &c.
Q. E. D.
a 3. 1.
To biscct a given rectilineal angle, that is, to divide it into two equal angles.
Let BAC be the given rectilineal angle, it is required to bisect it.
Take any point D in AB, and from AC cut a off AE equal
PROP. X. PROB.
Let AB be the given straight line; it is required to divide
Describe a upon it an equilateral triangle ABC, and bisect the angle ACB by the straight line CD. AB is cut into two equal parts in the point D.
c 8. 1.
a l. 1. b 9. 1.