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, 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 DFE. A D For if the triangle ABC be applied to DEF, so that the point ▲ may be on D, and the straight line AB upon DE; 1. The point B shall coincide with the point E, because AB is equal to DE. And AB coinciding with DE, because the angle BAC is equal to the angle EDF; wherefore also 3. 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 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 (Ax. 10.) is impossible; therefore 4. The base BC shall coincide with the base EF, and (Ax. 8.) be equal to it. Wherefore also 5. 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. 6. The angle ABC shall be equal 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.THEOREM. 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 (I. 3.) to AF, the less, and join FC, GB. Because AF is equal (Constr.) to 4G, and AB (Hypothesis) 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; therefore (I. 4.) 1. The base FC is equal to the base GB, and the triangle 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 the equal sides are opposite; viz. 2. The angle ACF is equal 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, (Ax. 3.) 3. 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 (I. 4.) 4. The triangles BFC, CGB, are equal, and their remaining angles are equal, each to each, to which the equal sides are opposite; therefore 5. 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; therefore 6. The remaining angle ABC is 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.-THEOREM. 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 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 (I. 3.) 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, 1. The two sides DB, BC, are equal to the two AC, CB, each to each; and the angle DBC is equal (Hyp.) to the angle ACB; therefore, (I. 4.) 2. 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. Therefore 3. 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.-THEOREM. 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. 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 sides 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, (I. 5.) 1. The angle ACD is equal to the angle ADC. But the angle ACD is greater than the angle BCD (Ax. 9.); therefore also 2. The angle ADC is greater than BCD; much more then 3. The angle BDC is greater than the angle BCD. Again, because CB is equal to DB, (I. 5.) 4. The angle BDC is equal to the angle BCD; but it has been demonstrated 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; therefore because AC is equal to AD in the triangle ACD, (I. 5.) the angles upon the other side of the base CD, namely, 1. The angles ECD, FDC, are equal to one another; but the angle ECD is greater than the angle BCD; wherefore likewise The angle FDC is greater than BCD; 2. much more then 3. The angle BDC is greater than the angle BCD. Again, because CB is equal to DB, (I. 5.) 4. The angle BDC is equal 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, upon the same base, and on the same side of it, &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 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 is equal to the angle EDF. A D G For if the triangle ABC be applied to DEF, so that the point B be on E, and the straight line BC upon EF; 1. The point C shall coincide with the point F, because BC is equal to EF; therefore BC coinciding with EF, 2. BA and AC shall coincide with ED and DF; for, if the base BC coincide 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, upon the same base EF, and upon the same side of it, there can be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise their sides terminated in the other extremity. But this is impossible (I. 7.); therefore, if the base BC coincide with the base EF, the sides BA, AC, cannot but coincide with the sides ED, DF; wherefore likewise 3. The angle BAC coincides with the angle EDF, and is equal (Ax. 8.) to it. Therefore if two triangles, &c. Q.E.D. PROP. IX.-PROBLEM. To bisect 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 Din AB, and from AC cut (I. 3.) off AE equal to AD; join DE, and upon it, on the side remote from 4, describe (I. 1.) an equilateral triangle DEF; then join AF; the straight line AF bisects the triangle BAC. Because AD is equal to AE, and AF is common to the two triangles DAF, EAF; 1. The two sides DA, AF, are equal to the two sides EA, AF, each to each; |