BA. But because neither is greater than the other, they are equal. And by parity of reasoning, the like may be proved in every other triangle whereof two angles are equal to one another. Wherefore, universally, if two angles &c. Which was to be demonstrated. See Note. COR. Every equiangular triangle is also equilateral. For if ABC be a triangle of which all the angles are equal to one another, then because the angles A and B are equal, the side CB (by Prop. VI above) is equal to the side CA. Again, because the angles C and B are equal, the side AB is equal *Interc. 1. to the side CA. Wherefore the sides CB and AB are* equal to one another. And since CB and AB are also equal to CA, the sides CB, CA, AB are all equal to one another. A B PROPOSITION VII. THEOREM. Upon the same given straight line and on the same side of it, there cannot be tmo triangles that have their sides which are terminated in one extremity of it equal to one another, and also those which are terminated in the other See Note. extremity. For if this be disputed, let it be assumed that there are two triangles ACB, ADB, upon the same given straight line AB and on the same side of it, which have their sides AC, AD, that are terminated in the extremity A, equal to one another, and also their sides BC, BD, that are terminated in the other extremity B. First Case ; where the vertex of each of B 1 1.5. ist equal to ADC. But the angle ACD is greater than BCD; therefore the angle ADC (which is equal [Interc. 1. to ACD) is alsof greater than BCD; still more is the angle Cor. 2. BDC (which is greater than ADC) greater than BCD. Again, because BC is equal to BD, the angle BDC must be equal to BCD; but it has been shown to be greater; and that it should E *I.Nom. 26. be both equal and greater, is impossible. The assumption*, there fore, that AC and AD are equal to one another, and also BC and BD equal to one another, cannot be true. Second Case; where one of the vertices, as D, is within the other triangle ACB. Join CD; А B 11.5. Cor.1. side of CD are equal to one another. But the angle ECD is greater than BCD; therefore the angle FDC (which is equal * INTERC. 1. to ECD) is also* greater than BCD; still more is the angle Cor. 2. BDC (which is greater than FDC) greater than BCD. Again, because BC is equal to BD, the angle BDC must be equal to BCD; but it has been shown to be greater; and that it should +1.Nom.26. be both equal and greater, is impossible. The assumptiont, there fore, that AC and AD are equal to one another, and also BC Third Case; where one of the vertices, as In this case BC and BD are not equal to one А B And by parity of reasoning, the like may, in any of the Cases, be proved of all other triangles under the same conditions. Wherefore, universally, upon the same given straight line &c. Which was to be demonstrated. Cor. If about two centres be described two circles that cut one another, their circumferences shall coincide only in two points, one on each side of the straight line that joins the centres. For if they coincided in more, then upon the given straight line which is between the centres and on the same side of it, might be two triangles having their sides which are terminated in one extremity of it equal to one another, and also those which are terminated in the other extremity. SCHOLIUM.–From this it follows that the two circles described in Prop. I of the First Book, could cut one another only in one point on each side of the straight line on which an equilateral triangle was required to be described; and consequently no more equilateral triangles than one could possibly be described on each side.-See Scholia to Prop. I and II of the First Book. F PROPOSITION VIII. two sides of the other respectively, and have also their third DG B CE For, let the triangle BAC be applied to the triangle EDF, so that "INTERC.12. Cor. 3. the point B may be on E, and the straight line BC on* EF. + Hyp. Because BC ist equal to EF, the point C will coincide with the point F. And, BC coinciding with EF, BA and CA will coincide with ED and FD. For, if BC coincides with EF, but the sides BA, CA do not coincide with the sides ED, FD, but have any different situation as EG and FG or otherwise ; then upon the same given straight line EF and on the same side of it, there will be two triangles which have their sides that are terminated in one extremity of it equal to one another, and also 11.7. their sides that are terminated in the other extremity; which ist impossible. Therefore the sides BA, CA cannot but coincide with the sides ED, FD; wherefore the angle BAC will coincide with the angle EDF, and be equal to it; and the whole triangle BAC will coincide with the whole triangle EDF, and be equal to it; and the remaining angles of the one will coincide with the remaining angles of the other respectively, and be equal to them, viz. the angle ABC to DEF, and ACB to DFE. And by parity of reasoning, the like may be proved of any other triangles under the same conditions. Wherefore, universally, if two triangles have two sides of the one equal to two sides of the other respectively, and have also their third sides equal; &c. Which was to be demonstrated. PROPOSITION IX. See Note. PROBLEM. To bisect a given straight line. [That is, to divide it into two equal parts.] B В D E E Let AB be the given straight line. It is reg quired to bisect it. • 1. 1. Describe* upon it an equilateral triangle ACB, +INTERC.12. and prolongt CA, CB, to D and E. Upon AB Cor. 6. and on the other side of it, describe another equilateral triangle AFB. Because the angles 11.5. Cor.1. BAD and ABE are equal to one another, and *1.5. Cor.2. the anles BAF and ABF are* equal to one another, and AF and BF meet in F, the angles BAF and ABF are respectively less than BAD and ABE; for if they were respectively equal, AF and BF would never meet on the side of F; still less if they were greater; and because they are neither equal nor greater, they are less; and because they are less, the point F which is in both AF and BF lies between the straight lines AD and BE. Join CF, and let CF cut AB in G. AB is bisected in G. + Constr. Because in the triangles CAF, CBF, CA ist equal to CB, and Constr. CF common, and AF equal* to BF; the angle ACF is* equal to *1.8. BCF. And because in the triangles ACG, BCG, GA ist equal + Constr. to CB, and CG is common, and the angle ACG (or ACF) has been shown to be equal to BCG (or BCF); the third side AG I 1. 4. ist equal to the third side BG; or AB is bisected in G. Which was to be done. And by parity of reasoning, in like manner may every other given straight line be bisected. F PROPOSITION X. straight line, from an assigned point in the same. Take any point D in AC, and make* A D C E B * 1. 3. + 1. 1. "Interc. 9. DFE, and join* CF. The straight line CF drawn from the Cor. assigned point C, is at right angles to AB. . + Constr. Because CD ist equal to CE and CF common to the two triangles DCF and ECF, the two sides DC, CF are equal to the I Constr. two EC, CF respectively. Also the third side DF ist equal to * I. 8. the third side EF; therefore the angle DCF is* equal to ECF; and they are adjacent angles. But when the adjacent angles which one straight line makes with another straight line are #I. Nom.37. equal to one another, each of them ist a right angle; therefore each of the angles DCF, ECF, is a right angle. Wherefore, from the point C in the straight line AB, CF has been drawii at right angles to AB. Which was to be done. And by parity of reasoning, the like may be done in every other instance. PROPOSITION XI. DI F E • Hyp: Cor.3. See Note. THEOREM.-All right angles are equal to one another. Let ABC, DEF be two right angles. They are equal to one another. (Interc.12. Prolong# CB to G, and FE C B G H Cor. 6. to H; and because ABC is* a right angle, ABC and ABG aret equal +1. Nom.37. to one another; and in like manner the angles DEF and DEH are equal to one another. Let now the straight line CBG be applied to the straight line FEH, so that the point B shall be Interc.12. on E, and the straight line CBG ont FEH; and let the whole figure CBGA be* turned round the straight line CG, till the * INTERC.9. Nom. point A be placed in the same plane in which are the points F, H, and D. If then BA does not coincide with ED, let it be assumed that it falls on the side of it which is towards H, as El. But because the angle IEF is equal to ABC, and IEH to ABG, and ABC TINTERC. 1. and ABG are equal to one another, IEF and IEH must bet Cor. 3. equal to one another. Wherefore the angles DEF and DEH must be equal to one another, and also the angles IEF and IEH equal to one another; which is impossible. For if DEF be equal to DEH, then because DEH is greater than IEH, INTERC. 1. DEF must be greater than IEH. The assumption*, therefore, Cor. 2 *].Nom.26. which involves the impossible consequence, cannot be true; or |