and B PROP. IV. THEOREM. to two sides of the other, each to each ; and Let ABC, DEF be two A CE F 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 + Hyp. with the point E, because AB is equal t to DE: and AB coinciding with DE, AC shall coincide + Hyp. with DF, because the angle BAC is equal t to the angle EDF; wherefore also the point C shall coincide with the point F, because the + Hyp. straight line AC + is equal to DF: but the point B was proved to coincide 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 coin cide with the base EF, two straight lines would 10 Ax. inclose a space, which is impossible.* Therefore the base BC coincides with the base EF, and 18 Ax. therefore is equalt to it. Wherefore the whole triangle ABC coincides with the whole triangle DEF, and is equal to it; and the other angles of the one coincide with the remaining angles of the other, and are 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 shall 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. ure equal to one another; and if the 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, and *3. 1. join FC, GB. Because AF is equal to +AG, and AB tofAC, Const. the two sides FA, AC are equal to the two GA, 6 Hyp. AB, each to each; and they contain the angle FAG common to the two triangles AFC, AGB; therefore the A base FC is equal* to the base GB, and the triangle AFC to the triangle AGB; and the remaining angles of the one are В, equal* to the remaining angles F G *4.1, of the other, each to each, to which the equal sides are oppo- D E site; viz. the angle ACF to the *4. 1. GB; * 4. 1. 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 *3 Ax. equal ; the remainder BF is equal * to the remainder CG; and FC was proved to be equal to therefore the two sides BF, FC are equal to the two CG,GB,each to each; and the angle BFC was proved to be equal to theangle CGB, and the base BC is common to the two triangles BFC, CGB; wherefore these triangles are equal*, 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; therefore the 73 Ax. remaining angle ABC is equaltto 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 triangle be egual 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. 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: therefore, because in the triangles DBC, ACB, DB is equal to AC, and BC com mon to both, 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; therefore the fIlyp. base DC is equal to the base AB, and the triangle DBC is equal to the triangle * ACB, the less to the D 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. B * *4.1. с Cor.—Hence every equiangular triangle is also equilateral. PROP. VII. THEOR. 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, upon the same base AB, and upon the same side of it, let there be two triangles ACB, ADB, which have their sides CA, DĂ terminated in the extremity A of the base, equal to one another, and likewise their sides, CB, DB, that are CD 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 t to A R + Hyp. AD, the angle ACD is equal* to the angle ADC: But the angle ACD is greater t +9 Ax. than the angle BCD; therefore the angle ADC is greater also than BCD; much more then is the angle BDC greater than the angle BCD. Again, because CB is equalt to DB, the angle + Hyp BDC is equal * to the angle BCD); but it has *5.1 been demonstrated to be greater than it; which is impossible. *5. 1. * с *5. 1. * * *5. 1. But if one of the vertices, as D, be within the other triangle ACB; produce AC, AD to E, F. + Hyp. therefore, because AC is equalt to AD in the triangle ACD, the E F equal * to one another: but the 19 Ax. angle ECD is greater t than the angle BCD: wherefore the angle Á than the angle BCD. Again, because CB is Hyp. equalt to DB, 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 nodemonstration. Therefore, 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. to two sides of the other, each to each, and have 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 A D G E |