It is required to cut off from AB the greater, a part equal to C, the From the point A draw the straight line AD equal to C; (1. 2) and from the centre A, at the distance AD, describe the circle DEF. (post. 3.) Then AE shall be equal to C. DEMONSTRATION Because A is the centre of the circle DEF, therefore AE is equal to AD; (def. 15) but the straight line C is likewise equal to AD; (constr.) therefore AE and C are each of them equal to AD; wherefore the straight line AE is equal to C. (ax. 1.) And therefore from AB, the greater of two straight lines, a part AE has been cut off equal to C, the less. Which was to be done. PROP. IV.-THEOREM 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 these sides equal to each other: then they shall likewise 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. (References Ax. 8, 10.) 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. Then 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. For, if the triangle ABC be applied to the triangle DEF. F so that the point A may be on D, and the straight line AB upon DE; because AB is equal to DE, therefore the point B shall coincide with the point E; and AB coinciding with DE, and the angle BAC being equal to the angle EDF, (hyp.) therefore AC shall coincide with DF; and because AC is equal to DF, wherefore also the point C shall coincide with the point F. 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 do not coincide with the base EF, two straight lines would inclose a space, which is impossible. (ax. 10.) 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 the angle DFE. Therefore, if two triangles have two sides of the one equal to two sides of the other, &c. Which was to be demonstrated. PROP. V.-THEOREM. The angles at the base of an isosceles triangle are equal to another; one 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. Then the angle ABC shall be equal to the angle ACB, and the angle CBD to the angle BCE. and from AE, the greater, cut off AG equal to AF, the less, (1.3) and join FC, GB. DEMONSTRATION Because AF is equal to AG; (constr.) and AB to AC; (hypoth.) 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; To assist the learner, the figure may be drawn so as to exhibit to the eye the triangles of which it is composed, in the following manner : 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 are equal to the remaining angles of the other, each to each, to which the equal sides are opposite, viz. the angle ACF to the angle ABG, and the angle AFC to the angle AGB. (1.4.) And because the whole AF, is equal to the whole AG, of which the parts AB, AC, are equal; the remainder BF shall be equal to the remainder CG; (ax. 3.) 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 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 is equal to the angle CBG. (1. 4.) And, since it has been demonstrated, that the whole angle ABG is equal to the whole angle ACF, the parts of which, the angles CBG, BCF, are equal; therefore 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 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. Cor. Hence every equilateral triangle is also equiangular. Q.E.L. PROP. VI.- THEOREM. If two angles of a triangle be equal to each other; then the sides also which subtend, or are opposite to, the equal angles shall be equal to one another. (References-Prop. 1. 3, 4.) Let the triangle ABC have the angle ABC equal to the angle ACB. Then the side AB shall be equal to the side AC. B CONSTRUCTION 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, (1. 3.) and join DC. DEMONSTRATION Then in the triangles DBC, ACB, because DB is assumed to be equal to AC, and BC is common to both, the two sides DB, BC, must be equal to the two AC, CB, each to each; and the angle DBC is equal to the angle ACB; (hyp.) therefore the base DC must be equal to the base AB, and the triangle DBC to the triangle ACB, (1. 4) the less equal to the greater, which is absurd. Therefore AB is not unequal to AC, that is AB is equal to AC. Wherefore, if two angles, &c. Cor. Hence every equiangular triangle is also equilaterai. Q.E.D. 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 each other, and likewise those which are terminated in the other extremity. (References Prop. 1. 5; ax. 9.) 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. |