The sign > signifies "is greater than," and < "is less than." + expresses addition ; thus AB + BC is the line whose length is the sum of the lengths of AB and BC. expresses subtraction; thus AB BC is the excess of the length of the line AB'above that of BC. the upon the straight line AB. Proposition 1.-Problem. From centres A and B, and radius=AB, describe circles, B To describe an equilateral triangle on a given finite straight line. Let AB be the given straight line. CONSTRUCTION.- From the centre A, at the distance AB, describe the circle BCD (Post. 3). From the centre B, at the distance BA, describe the circle ACE (Post. 3). From the point C, in which the circles cut one another, draw the straight lines CA, CB to the points A and B (Post. 1). Then ABC shall be an equilateral triangle. Because the point B is the centre of the circle ACE, BC is equal to BA (Def. 15). Therefore AC and BC are each of them equal to AB. But things which are equal to the same thing are equal to one another. Therefore AC is equal to BC (Ax. 1). Therefore AB, BC, and CA are equal to one another. Therefore the triangle ABC is equilateral, and it is described on the given straight line AB. Which was to be done. AC=AB. BC=AB. AC and BC each=AB. AC=BC. ..AB=BC =CA. B as centre. H D Das centre. A C с IB Proposition 2.- Problem. 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. CONSTRUCTION.- From the point A to B draw the straight Draw AB. line AB (Post. 1). Upon AB describe the equilateral triangle DAB (Book I., A DAB eProp. 1). quilateral. Produce the straight lines DA, DB, to E and F (Post. 2). From the centre B, at the dis K tance BC, describe the circle CGH, meeting DF in G (Post. 3). From the centre D, at the distance DG, describe the circle GKL, meeting DE in L (Post. 3). Then AL shall be equal to BC. BC=BG. the centre of the circle CGH, BC is equal to BG (Def. 15). Because the point D is the centre of the circle GKL, DL DL=DG. is equal to DG (Def. 15). But DA, DB, parts of them, are equal (Construction). DA=DB. Therefore the remainder AL is equal to the remainder BG AL = BG (Ax. 3). But it has been shown that BC equal to BG. Therefore from the given point A a straight line AL has :: AL = been drawn equal to the given straight line BC. Which was to be done. Proposition 3.- 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, of which AB is the greater. .. AL and BC each BC, Make AD =C. A as centre and radius AD. A B It is required to cut off from AB, the greater, a part equal to C, the less. CONSTRUCTION.–From the point A draw the straight line AD equal to C (I. 2). From the centre A, at the distance AD, describe the circle DEF, cutting Then AE shall be equal to C. But C is also equal to AD (Construction). Therefore AE and C are each of them equal to AD. AE=C. Therefore AE is equal to C (Ax. 1). Therefore, from AB, the greater of two given straight lines, a part AE has been cut off, equal to C, the less. Q. E. F.* AD=C. AE and C each=AD. Proposition 4.-Theorem. If two triangles have two sides of the one equal to two sides of the other, each to each, and have also the angles contained by those sides equal to one another : they shall 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. Or, If two sides and the contained angle of one triangle be re8pectively equal to those of another, the triangles are equal in every respect. Let ABC, DEF be two triangles which have each to each, viz., AB equal to And the angle BAC equal to the angle EDF:—then The base BC shall be equal to the base EF; The triangle ABC shall be equal to the triangle DEF; *Q. E. F. is an abbreviation for quod erat faciendum, that is “ which was to be done." AB=DE. AC=DF. A D 4 BAC = 4 EDF. B E A ABC A 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 the angle DFE. PROOF.-For if the triangle ABC be applied to (or placed Suppose upon) the triangle DEF, put upon So that the point A may be on the point D, and the straight line AB on the straight line DE, The point B shall coincide with the point E, because AB is equal to DE (Hypothesis). And AB coinciding with DE, AC shall coincide with DF, because the angle BAC is equal to the angle EDF (Hyp.). Therefore also the point C shall coincide with the point F, because the straight line AC is equal to DF (Hyp.). But the point B was proved to coincide with the point E. 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 enclose a space, which is impossible (Ax. 10). Therefore the base BC coincides with the base EF, and is BC=EF. therefore equal to it (Ax. 8). Therefore the whole triangle ABC coincides with the whole ::A ABC triangle DEF, and is equal to it (Ax. 8). And the other angles of the one coincide with the remain- LABC = ing angles of the other, and are equal to them, viz., the angle 2 ACB = ABC to DEF, and the angle ACB to DFE. Therefore, if two triangles have, &c. (see Enunciation). Which was to be shown. = A DEF. DFE. Proposition 5.-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 also be equal. Let ABC be an isosceles triangle, of which the side AB is AB = AC. equal to the side AC. Let the straight lines AB, AC (the equal sides of the triangle), be produced to D and E. The angle ABC shall be equal to the angle ACB (angles at the base), D E ..FC=GB and A AFC =A AGB And the angle CBD shall be equal to the angle BCE (angles upon the other side of the base). CONSTRUCTION.— In BD take any point F. AG=AF, From AE, the greater, cut off AG, equal to AF, the less (I. 3). Join FC, GB. PROOF.—Because AF is equal to AG (Construction), and AB is equal to AC (Нур.), FA, AC re Therefore the two sides FA, AC are spectively =GA, AB, equal to the two sides GA, AB, each to each; Therefore the base FC is equal to the base GB (I. 4); And the remaining angles of the one are equal to the 4 ACF = remaining angles of the other, each to each, to which the 2 AFC = equal sides are opposite, viz., the angle ACF to the angle ABG, and the angle AFC to the angle AGB (I. 4). And because the whole AF is equal to the whole AG, of which the parts AB, AC, are equal (Hyp.), The remainder BF is equal to the remainder CG (Ax. 3). Therefore the two sides BF, FC are equal to the two sides And the angle BFC was proved equal to the angle CGB; Therefore the triangles BFC, CGB are equal; and their other angles are equal, each to each, to which the equal sides are opposite (I. 4). 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 angle ACF, and that the parts of these, the angles CBG, BCF, are also equal, Therefore the remaining angle ABC is equal to the remaining angle ACB (Ax. 3), Which are the angles at the base of the triangle ABC, 2 AGB, BF=CG. ... FBC = L GCB. L BCF = 2 CBG. ..L. ABC |