B as centre. K H Das centre A BC=BG. 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 A B describe the equilateral triangle DAB (Book I., A DAB Prop. 1). quilateral. Produce the straight lines DA, DB, to E and F (Post. 2). From the centre B, at the distance 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. PROOF.-Because the point B is 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 BGAL = (Ax. 3). But it has been shown that BC is equal to BG. But things which are equal to the same thing are equal to BG. cne another, therefore AL is equal to BC (Ax. 1). Therefore from the given point A a straight line AL has 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 .. AL BC. Make AD A as centre and radius AD. 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. PROOF.—Because the point A is the centre of the circle AE=AD. DEF, AE is equal to AD (Def. 15). But C is also equal to AD (Construction). 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. 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 respectively 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 <BAC EDF. B A ABC 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, So that the point A may be on the point D, and the A DEF. 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 ACE = ABC to DEF, and the angle ACB to DFE. Therefore, if two triangles have, &c. (see Enunciation). Thich was to be shown. = A DEF. . 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), And the angle CBD shall be equal to the anglo 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 Hyp.), FA, AC rc Therefore the two sides FA, AC aro spectively GA, AB. equal to the two sides GA, AB, each to D E each ; .. FC=C3 and AFC =AAGB. L ACT = LABG. LANCE LAGE, BF=CC. And they contain the angle FAG, common to the two triangles AFC, AGB. Therefore the base FC is equal to the base GB (T. 4); And the remaining angles of the one are equal to the remaining angles of the other, cach to each, to which the cqual 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 Ar 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 EF, FC are equal to the two sides And the anglo BFC was proved cqual to the angle CGB; Therefore the triangles BFC, CGB are equal; anıl 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 anglo ABG is equal to the whole angle ACF, and that the parts of these, the angles CBG, BCF, are also cqual, Therefore the remaining angle ABC is equal to the remaining angle ACB (Ax. 3), Which are the angles at the base of the trianglo ABC, And it has been proved that the angle FBC is equal to the angle GCB (Dem. 11), Which are the angles upon the other side of the base, Therefore the angles at the base, &c. (see Enunciation). which was to be shown. COROLLARY.-Henco every equilateral triangle is also equiangular. AB >AC. DB = A. Proposition 6.-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 shall be equal to the side AC. For if AB be not equal to AC, one of them is greater Suppose than the other. Let AB be the greater. CONSTRUCTION.-From AB, the greater, cut off a part DB, Mako equal to AC, the less (I. 3). Join DC. Proor.— Because in the triangles DBC, ACB, DB is equal to AC, and BC is common to both, Therefore the two sides DB, BC are equal to the two sides AC, CB, each to each ; And the angle DBC is equal to the anglo ACB (Hyp) Therefore the base DC is equal to the base AB (I. 4). And the triangle DBC is equal to the triangle ACB (I. 4), the less to the greater, which is absurd. Therefore AB is not unequal to AC, that is, it is equal to it. COROLLARY.-Hence every equiangular triangle is also cquilateral. B CA DBC A ACB. Q. E. D. is an abbreviation for quod erat demonstrandum, that is, "which was to be shown or proved," |