For if the triangle ABC be applied to the triangle DEF, so that the point A may be on the point D, and the straight line AB on the straight line DE, the point B will coincide with the point E, because AB is equal to DE. [Hyp. And, AB coinciding with DE, AC will fall on DF, because the angle BAĆ is equal to the angle EDF. AA [Hypothesis. Therefore also the point C will coincide with the point F, because AC is equal to DF. [Hypothesis. But the point B was shewn to coincide with the point E, therefore the base BC will coincide with the base EF; because, B coinciding with E and C with F, if the base BC does not coincide with the base EF, two straight lines will enclose a space; which is impossible. [Axiom 10. Therefore the base BC coincides with the base EF, and is equal to it. [Axiom 8. Therefore the whole triangle ABC coincides with the whole triangle DEF, and is equal to it. [Axiom 8. And the other angles of the one coincide with the other angles of the other, and are equal to them, namely, the angle ABC to the angle DEF, and the angle ACB to the angle DFE. Wherefore, if two triangles &c. Q.E.D. PROPOSITION 5. THEOREM. The angles at the base of an isosceles triangle are equal to one another; and if the equal sides be produced the angles on the other side of the base shall be equal to one another. Let ABC be an isosceles triangle, having the side AB equal to the side 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, [I.3. and join FC, GB. Because A Fis equal to AG, [Constr. and AB to AC, [Hypothesis. the two sides FA, ACare equal to the two sides GA, AB, each to each; and they contain the angle FAG common to the two triangles AFC, AGB; therefore the base FC is equal to the base GB, and the triangle AFC to the triangle AGB, and the remaining angles of the one to the remaining angles of the other, each to each, to which the equal sides are opposite, B namely the angle ACF to the angle ABG, and the angle AFC to the angle AGB. [I. 4. And because the whole AF is equal to the whole AG, of which the parts AB, AC are equal, [Hypothesis. the remainder BF is equal to the remainder CG. [Axiom 3. And FC was shewn to be equal to GB; therefore the two sides BF, FC are equal to the two sides CG, GB, each to each; and the angle BFC was shewn to be equal to the angle CGB; therefore the triangles BFC, CGB are equal, and their other angles are equal, each to each, to which the equal sides are opposite, namely the angle FBC to the angle GCB, and the angle BCF to the angle CBG. [I. 4. And since it has been shewn that the whole angle ABG is equal to the whole angle ACF, and that the parts of these, the angles CBG, BCF are also equal; therefore the remaining angle ABC is equal to the remaining angle ACB, which are the angles at the base of the triangle ABC. [Axiom 3. And it has also been shewn that the angle FBC is equal to the angle GCB, which are the angles on the other side of the base. Corollary. Hence every equilateral triangle is also equiangular. PROPOSITION 6. THEOREM. If two angles of a triangle be equal to one another, the sides also which subtend, or are oppo site 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 AC shall be equal to the side AB. For if AC be not equal to AB, one of them must be 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, and BC is common to both, B [I. 3. [Construction. the two sides DB, BC are equal to the two sides AC, CB, each to each; and the angle DBC is equal to the angle ACB; [Hypothesis. therefore the base DC is equal to the base AB, and the triangle DBC is equal to the triangle ACB, the less to the greater; which is absurd. [I. 4. [Axiom 9. Therefore AB is not unequal to AC, that is, it is equal to it. Wherefore, if two angles &c. Q.E.D. Corollary. Hence every equiangular triangle is also equilateral. On the same base, and on the same side of it, there cannot be two triangles having their sides which are terminated at one extremity of the base equal to one another, and likewise those which are terminated at the other extremity. If it be possible, on the same base AB, and on the same side of it, let there be two triangles ACB, ADB, having their sides CA, DA, A which are terminated at the extremity A of the base, equal to one another, and likewise their sides CB, DB, which are terminated at B. Join CD. In the case in which the vertex of each triangle is without the other triangle ; because AC is equal to AD, [Hypothesis. the angle ACD is equal to the angle ADC. [I. 5. But the angle ACD is greater than the angle BCD, [4x. 9. therefore the angle ADC is also 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 BDC is equal to the angle BCD. [Hypothesis. [I. 5. But it has been shewn to be greater; which is impossible. Then because AC is equal to AD, in the triangle ACD, [Hyp. the angles ECD, FDC, on the other side of the base CD, are equal to one another. [I. 5. But the angle ECD is greater than the angle BCD, A [Axiom 9. therefore the angle FDC is also 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 BDC is equal to the angle BCD. [Hypothesis. [I. 5. But it has been shewn to be greater; which is impossible. The case in which the vertex of one triangle is on a side of the other needs no demonstration. Wherefore, on the same base &c. Q.E.D. 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 which is 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, namely AB to DE, and AC to DF, and also the base BỞ equal to the base EF: the angle BÁC shall be equal to the angle EDF. For if the triangle ABC be applied to the triangle DEF, so that the point B may be on the point E, and the straight line BC on the straight line EF, the point C will also coincide with the point F, because BC is equal to EF. [Hyp. Therefore, BC coinciding with EF, BA and AC will coincide with ED and DF. For if the base BC coincides with the base EF, but the sides BA, CA do not coincide with the sides ED, FD, but have a different situation as EG, FG; then on the same base and on the same side of it there will be two triangles having their sides which are terminated at one extremity of the base equal to one another, and likewise their sides which are terminated at the other extremity. But this is impossible. [I. 7. Therefore since the base BC coincides with the base EF, the sides BA, AC must coincide with the sides ED, DF. Therefore also the angle BAC coincides with the angle EDF, and is equal to it. [Axiom 8. Wherefore, if two triangles &c. Q.E.D. PROPOSITION 9. PROBLEM. To bisect a given rectilineal angle, that is to divide it into two equal angles. |