From the centre A, at the distance AB, describe the circle BCD, and from the centre B, at the distance BA, describe the circle ACE; and from C, where the circles cut one another, join CA, CB; ABC is an equilateral triangle. BE Because A is the centre of BCD, AC is equal to AB; and because B is the centre of ACE, BC is equal to BA, but it has been proved that CA is equal to AB; therefore CA, CB, are each of them equal to AB; but things which are equal to the same are equal to one another; therefore CA is equal to CB; wherefore CA, AB, BC, are equal to one another and the triangle ABC is equilateral, and is described on the given straight line AB. ; II.—From a given point to draw a straight line equal to a given straight line. Let A be the given point, and BC the given straight line; it is required to draw from the point A a straight line equal to BC. Join AB, and on it describe the equilateral triangle DAB, and produce DA, DB, to E and F; from the centre B, at the distance BC, describe the circle CGH, and from the centre D, at the distance DG, describr the circle GKL. Because B is the centre of CGH, BC is equal to BG; and because D is the centre of GKL, DL is equal to DG; and DA, DB, parts of them, are equal; therefore the remainder AL, is equal to the remainder BG; but it has been shown that BC is equal to BG; wherefore AL and BC are each of them equal to BG; and things that K H D A B are equal to the same are equal to one another; therefore AL is equal to BC. Wherefore from the given point A a straight line AL has been drawn equal to the given straight line BC. III. From the greater of two given straight lines to cut off a part equal to the less. Let AB and C be the two given straight lines, whereof AB is the greater. It is required to cut off from AB a part equal to C, the less. From A draw AD equal to C; and EB from the centre A, and at the distance AD, describe the circle DEF; and because A is the centre of DEF, AE shall be equal to AD; but C is likewise equal to AD; whence AE and C are each of them equal to AD; wherefore the straight line AE is equal to C, and from AB, the greater of two straight lines, a part AE has been cut off equal to C the less. IV.-If two triangles have two sides of the one equal to two sides of the other, each to each; and likewise the angles contained by those sides equal to one another; 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. Let ABC, DEF, be two triangles, which have the sides AB, AC, equal to 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 B D CE F 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 cach, viz., ABC to DEF, and ACB to DFE. 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 with the point E, because AB is equal to DE; and AB coinciding with DE, AC shall coincide with DF, because the angle BAC is equal to the angle EDF; wherefore also C shall coincide with F, because 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 is impossible. Therefore BC shall coincide with 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., ABC to DEF, and ACB to DFE. V.-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 AB is equal to AC, and let AB, AC be produced to D and E, the angle ABC shall be equal to ACB, and the angle CBD to BCE. In BD take any point F, and from AE the greater, cut off AG equal to AF the less, and join FC, GB. Because AF is equal to AG, and AB 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 the F Ꭰ B E 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., ACF to ABG, and AFC to 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 BG; therefore the two BF, FC, are equal to the two CG, GB, each to each, and the angle BFC is equal to CGB, and BC is common to the two triangles BFC, CGB; wherefore the triangles are equal and their remaining angles each to each, to which the equal sides are opposite; therefore FBC is equal to GCB, and BCF to CBG; and, since it has been demonstrated that the whole angle ABG is equal to the whole ACF; the parts of which, CBG, BCF, are also equal, the remaining angle ABC is therefore equal to the remaining angle ACB; which are the angles at the base of the triangle ABC; and it also 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. COROLLARY.-Hence every equilateral triangle is also equiangular. VI.-If two angles of a triangle be eqaul 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; let AB be the greater, and from it cut off D BD 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, the two sides DB, BC, are equal to the two AC, CB, each to each; and the angle DBC is equal to ACB; therefore DC is equal to AB, 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. B COROLLARY.-Hence every equiangular triangle is also equilateral. VII.-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, because AC is equal to AD, the angle ACD is equal to ADC; but the angle ACD is greater than BCD; A C D therefore the angle ADC is greater also than BCD; much more then is the angle BDC greater than BCD. Again, because CB is equal to DB, the angle BDC is equal to BCD; but it has been demonstrated to be greater, which is impossible. But if one vertex, as D, be within the other triangle ACB, produce AC, AD, to E and F; therefore, because AC is equal to AD in the triangle ACD, the angles ECD, FDC, upon the other side of the base CD are |