equal to AD (Constr.) ; whence AE and C are each of them equal to AD: Wherefore AE is equal to C, and from AB, the greater of two given straight lines, a part AE has been cut off equal to C the less. Q.E.F. PROP. IV, THEOREM. If two triangles have two sides of the one equal to two sides of the other, each to each, and have also the angles contained by those sides equal to one another, they shall also 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 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 shall be equal, each to each, to which the equal sides are opposite, viz. the angle ABC to the angle DEF, and the angle ACB to the angle DFE. For, if the triangle ABC be applied to the triangle 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 (Hyp.): and AB coinciding with DE, AC shall coincide with DF, because the angle BAC is equal to the angle EDF (Hyp.) : therefore also the point C shall coincide with the point F, because AC is equal to DF (Hyp.): But the point B coincides with the point E, therefore the B base BC shall coincide with the base EF; because, B coinciding with E, and C with F, if BC does not coincide with EF, two straight lines would enclose a space—which is impossible (Ax. 10); therefore the base BC shall coincide with the base EF, and be equal to it (Ax. 8); and 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 other angles of the other, and be equal to them, each to each, viz. the angle ABC to the angle DEF, and the angle ACB to the angle DFE. Wherefore, If two triangles &c. Q.E.D. 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 (1. 3), and join FC, GB. Then, because AF is equal to AG, and AB to AC (Hyp.), 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. fore (1. 4) the base FC is equal to the base D GB, and the triangle AFC to the triangle AGB, and the remaining angles of the one to the A B с F 1 remaining angles of the other, each to cach, to which the equal sides are opposite, viz. the angle 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 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-therefore (1. 4) (the base BC being here common to the two triangles BFC, CGB) the two triangles are equal, and their other angles are equal, each to each, to which the equal sides are opposite, viz. the angle FBC 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 angle ACF, parts of which, the angles CBG, BCF, are also equal, the remaining angle ABC is therefore equal to the remaining angle ACB (Ax. 3), which are the angles at the base of the triangle ABC: And it has been proved that the angle FBC is equal to the angle GCB, which are the angles upon the other side of the base. Wherefore, 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 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. B 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: Then, because in the triangles DBC, ACB, DB is equal to AC, and BC common to both triangles, the two sides DB, BC are equal to the two AC, CB, each to each, and the angle DBC is equal to the angle ACB (IIyp.)therefore the base DC is equal to the base AB (1. 4), and the triangle DBC is equal to the triangle ACB, the less to the greater—which 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. D PROP. VII. THEOR. Upon the same ase, and on the same side of it, there cannot be two triangles which have their sides terminated in one extremity of the base equal to one another, and also those 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 also their sides CD, DB, equal, 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 to the angle ADC (1. 5): But the angle ACD is greater than the angle BCD (Ax. 9); therefore also the angle ADC is greater than the angle BCD; much R more then is the angle BDC greater than the angle BCD: Again, because BC is equal to BD the angle BCD is equal to the angle BDC: but it has been proved to be greater than it—which is impossible. But, if one of the vertices, as D, be within the other triangle ACB, produce AC, AD to E, F. Then, because AC is equal to AD in the triangle ACD, the angles ECD, FDC upon the other side of the base CD are equal to one another (1.5): But the angle ECD is greater than the angle BCD; therefore also the angle FDC is greater than the angle BCD; much more then is the angle BDC greater than the angle BCD: Again, because BC is equal to BD, the angle BCD is equal to the angle BDC: but it has been proved to be greater than it—which is impossible. The case, in which the vertex of one triangle is upon a side of the other, needs no demonstration. Wherefore, Upon the same bas &c. 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 also 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 sides, equal to them, of the other. Let ABC, DEF be two triangles, having the two sides AB, AC equal to the two sides DE, DF, each to each, viz. AB to DE, and AC to DF, and also the base BC equal to the base EF : the angle BAC shall be equal to the angle EDF. |