PROPOSITION I.-PROBLEM. To describe an equilateral triangle upon a given finite straight line. Let AB be the given straight line; it is required to describe an equilateral triangle upon it. From the centre A, at the distance AB, describe (Postulate 3.) the circle BCD; and from the centre B, at the distance BA, describe the circle ACE; and from the point C, in which the circles cut one another, draw (Post. 1.) the straight lines CA, CB, to the points A, B; ABC shall be an equilateral triangle. E Because the point A is the centre of the circle BCD, (Definition 15.) 1. AC is equal to AB; and because the point B is the centre of the circle ACE, 2. 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 (Ax. 1.); therefore 3. CA is equal to CB. Wherefore CA, AB, BC, are equal to one another; and therefore 4. The triangle ABC is equilateral; and it is described upon the given straight line AB. Which was to be done. PROP II.-PROBLEM. From a given point to draw a straight line equal to a given straight line. Let 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. From the point A to B draw (Post. I.) the straight line AB; and upon it describe (I. 1.) the equilateral triangle DAB, and produce (Post. 2.) the straight lines DA, DB to E and F. From the centre B at the distance BC, describe (Post. 3.) the circle CGH, and from the centre D, at the distance DG, describe the circle GKL. AL shall be equal to BC. K Because the point B is the centre of the circle CGH, (Def. 15.) 1. BC is equal to BG; and because D is the centre of the circle GLK, 2. DL is equal to DG; and DA, DB, parts of them are equal (Construction); therefore, (Ax. 3.) 3. 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 are equal to the same are equal to one another; therefore 4. The straight line 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. Which was to be done. PROP. III.-PROBLEM. 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, the greater, a part equal to C, the less. From the point 4, draw (I. 2.) the straight line AD equal to C; and from the centre A, and at the distance AD, describe (Post. 3.) the circle DEF: AE shall be equal to C. D D B Because A is the centre of the circle DEF, (Def. 15.) 1. AE is equal to AD; but the straight line C is likewise equal to AD (Constr.); whence AE and Care each of them equal to AD; wherefore (Àx. 1.) 2. 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 the less. Which was to be done. 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; 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 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 B E 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; 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 4C 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 4C, 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 4G equal (I. 3.) to AF, the less, and join FC, GB. Because AF is equal (Constr.) to AG, 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. D B 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, |