PROPOSITION XXII. PROBLEM.—To describe a triangle of which the sides shall be equal to three given straight lines respectively ; provided always, that K к D -E are together greater than the third ; (for if they are not, it has been *I. 20.Cor. shown* that they cannot form a triangle). It is required to make a triangle, of which the sides shall be equal to A, B, C, respectively. + 1. 3. In a straight line DE, maket FG equal to any one of the straight lines as A; FH equal to another as B; and GI equal to Interc.13. C. About the centre F, with the radius FH, describes a circle ; Cor. 3. *Interc.13. and let it cut FE in the* point N. About the centre G, with the Cor. 5. radius GI, describe another circle ; and because FH and GI are tHyp. togethert greater than FG, this last circle will cut GD in a point (as M) nearer to D than is the point N. For if M and N should coincide, FH and GI must be together equal to FG; and if M should lie on the other side of N, they must be together less than Hyp. FG; neither of which is possible, for they are* greater. Also because FN (which is equal to FH) is* less than the sum of FG and GI, the point N will fall on the side of I which is towards D; + Hyp. and because GM (which is equal to GI) ist less than the sum of GF and FH) the point M will fall on the side of H which is towards E. Wherefore the circumference of each of the circles will pass through that of the other; and in passing from the outside of it to the inside, and from the inside to the outside again, 11. 7. Cor. will necessarily cut it in two pointst. Let one of these be K. Join KF, KG. The triangle FKG has its sides equal to the three straight lines A, B, C. *Interc.13. Because the point F is the centre of the circle HKL, FK is* Cor. 4. equal to FH. But FH ist equal to the straight line B; thereInterc. 1. fore FK ist equal to B. Again, because G is the centre of the Cor. 1. circle IKL, GK is equal to GI. But GI is equal to the straight line C; therefore GK is equal to C. Also FG is equal to A; therefore the three straight lines FG, FK, GK (which are the sides of the triangle FKG) are respectively equal to the three straight lines A, B, C. Which was to be done. • Hyp: + Constr. And by parity of reasoning, the like may be done in every other instance. с Cor. 1. In this manner may be described a triangle of which the sides shall be equal to those of a given triangle respectively. And if it is required that one particular side of the triangle shall be on an assigned straight line DE, and terminated at an assigned point F in it, and that another particular side shall likewise be terminated at F; this also may be done. For it may be effected by making FG equal to the side which is to be on DE, and FH equal to the other side which is to be terminated at F, and GI equal to the remaining side. Cor. 2. By the help of this, on an assigned straight line and at an assigned point in it, may be made an angle equal to a given angle. For if AB be the assigned straight line, A the point in it, and C the given angle; in the two straight lines between which is the angle C, take any points D, B E, and join DE; and (by Prop. XXII above) describe the triangle AFG, in which the side AF shall be equal to CD, and AG to CE, and FG to DE. Because AF, AG are equal to CD, CE respectively, and the third side FG is . 1.8. equal to the third side DE, the angle FAG will be* equal to the angle DCE. Cor. 3. Hence also on an assigned straight line and at an assigned point in it, may be described a rectilinear figure having its sides and angles respectively equal to those of a given rectilinear figure of any number of sides ; and the two rectilinear figures shall be equal. For the given rectilinear figure may be divided into triangles, by straight lines drawn from some of its angular points to others. And triangles having their sides equal to the sides of these respectively, may (by Cor. 1 above) be constructed in succession, and in the corresponding positions. And because in the corresponding triangles the angles will bet equal respectively, viz. those to which equal sides are opposite; by adding equal angles to equal, the Interc. 1. angles of one of the rectilinear figures will be equal to the angles Cor. 5. of the other respectively. And because the several triangles are equal respectively, the sums, which are the rectilinear figures, are equal. PROBLEM.—To draw a straight line perpendicular to an as signed straight line of unlimited length, from an assigned point See Note. without it. Let AB be the assigned straight line, which may be prolonged In AB take any point, as D; and join CD. If CDB is a to AB as required. But if CDB is not a right angle, then on the •1.22.Cor. 2. other side of AB make* an angle BDF equal to BDC, and maket tl. 3. DF equal to DC, and join CF. Because the equal angles CDB and FDB are either greater or less than right angles, and all 11.12.Cor.2. the angles at the point D aret together equal to four right angles ; DC and DF make an angle less than two right angles, either on one 113.Cor.3. side or the other. Wherefore CF will* form a triangle with them, on that side on which is the angle CDF that is less than the sum +1.9. of two right angles. Bisectt CF, in G; and join DG. Because Constr. in the triangles CDG, FDG, DC ist equal to DF, and CG to* Constr. FG, and DG is common; the triangles aret equal, and the angle +1.8. CDG is equal to FDG, and DGC to DGF, and because they are #I.Nom.37. adjacent angles they are right angles. And because the angle CDF has been shown to be bisected by the straight line DG, and Constr. was also* bisected by the straight line DB, DG and DB aret in + 1. 9. Cor. 11.13.Cor.2. one straight line; wherefore EB shall pass through G, and cutt 1. 12. CF. And because CGD and CGB are* together equal to two right angles, and CGD is a right angle, CGB is also a right angle, +I.Nom.37. and the straight line CG drawn from the point C ist perpendicular to AB. Which was to be done. And by parity of reasoning, the like may be done in every other instance. Cor. 1. Of all the straight lines that can be drawn to a straight line from a point without it, the perpendicular is the shortest. 11.Nom.48. For any other will be the hypotenuse of a right-angled triangle, "1. 19. Cor. and therefore* greater than the perpendicular. On Cor. I prece 77 ding, See Note. COR. 2. If two straight lines make an acute angle ACD; and from a point A in one of 1. 23. them, a straight line be drawn* perpendicular ! to the other; the perpendicular shall fall on that B C D For the perpendicular cannot fall upon the point C; because +Нур. . ACD ist not a right angle but an acute angle. Neither can it fall on the side of C on which is the obtuse angle, as for instance upon the point B; for since ACD (which is the exterior angle of the triangle ACB) is less than a right angle, still more is 11. 16. ABC (which is less than ACD) less than a right angle. Therefore, because it neither falls upon C, nor on the side of it towards B, it falls on the side towards D; that is, on which is the acute angle. COR. 3. If in any triangle a side AB be *1.17.Cor.3. taken which lies between* two acute angles, and from the opposite angular point a perpen,+ 1.23. dicular CD be drawnt to the side ; the per pendicular shall fall between the extremities of the side. For because BAC is an acute angle, CD (by Cor. 2 above) falls on the side of A which is towards B. And because ABC is an acute angle, CD falls on the side of B which is towards A. Therefore it falls between A and B. ! THEOREM.-If two triangles have two sides of the one, equal to two sides of the other respectively, but the angle between the two sides of the one triangle is greater than the angle between the two sides which are equal to them in the other triangle ; the remaining side of that which has the greater angle, shall be greater than the remaining side of the other. See Note. Again, the than the angle EDF. The remaining side BC is greater than Of the two sides AB, AC, let AB be one that is not greater than the other; and at the point A in the straight line *1.22.Cor.2. AB, make* the angle BAI equal to the angle EDF. Because AB is not greater than AC, the angle ACB is not-greater than + I. 19. ABC; for if it was greater, AB the side opposite to it would bet greater than AC the side opposite to the other. [ I. 16. exterior angle AHC ist greater than the interior opposite Therefore in the triangle AHC, the side AC which is opposite to • I. 19. the angle AHC the greater, is* greater than AH which is opposite to the angle ACH the less. Since, therefore, AH is less than + I. 3. AC, maket AG equal to AC, and the point G will be on the BG, GC, and the angle BAG ist equal to the angle EDF, BG isi 11.4. equal to EF. And because AC is* equal to AG, the angle *Constr. AGC ist equal to ACG. But the angle ACG is greater +1.5. than BCG; therefore the angle AGC (which is equal to ACG) INTERC. 1. isf greater than BCG. Still more is the angle BGC (which is Cor. 2. greater than AGC) greater than BCG. Wherefore in the triangle BGC, the side BC (which is opposite to the angle * I. 19. BGC the greater) is* greater than BG (which is opposite to the angle BCG the less). But BG has been shown to be +INTERC. 1. equal to EF; therefore BC (which is greater than BG) ist Cor. 2. greater than EF. And by parity of reasoning, the like may be proved in all other triangles under the same conditions. Wherefore, universally, if two triangles have two sides &c. Which was to be demonstrated. |