LOCUS. 



18. If triangles be infcribed in a given fegment of a 

 circle, and from the vertex on either fide (produced if ne- 

 celTary) there be taken, either way, a right line always in 

 a conflant ratio to either of the fides, or to their fum, or 

 difference, the loci of the points fo defcribed will be 

 circles. 



The above contain many of the moft fimple cafes of 

 geometrical loci ; and we will now (hew the application of 

 them to the conilru£lion of certain geometrical problems. 



Of the Conflruaion of geometrltal Prablimt. 



Pkob. I. 



Having given the bafe, perpendicular, and vertical angle 

 of a plane triangle ; it is required to conftruft it. Fig. 7. 



Analjfu — Suppofe the thinj; done, and let ABC repre- 

 fent the propofed triangle, of which the bafe A B, tlie per- 

 pendicular C D, and vertical angle A C B, are given ; 

 then it is obvious, in the iirft place, that the locus of the 

 vertex will be the right line C F, drawn parallel to A B, 

 at the given perpendicular diftance. Alfo, fince the angle 

 A C B is given, the locus of the vertex will be in the circum- 

 ference of the circle A C B, defcribed upon A B, capable 

 of containing the given angle A C B ; and confequentiy, at 

 either point where the line and circle interfedl each other, 

 ■will be the vertex of the triangle required. 



ConflruB'wn. — On the given bafe A B defcribe a circle 

 that (hall contain the given vertical angle. And parallel to 

 A B, and at a didance equal to the given perpendicular, 

 draw the nght line F C E ; join A C, C B, fo is A C B 

 the triangle required. For, the point C being in the feg- 

 ment A C B, the angle A C B is equal to the given ver- 

 tical angle ; and being alfo in the line F C E, the perpen- 

 dicular C D is equal to the given perpendicular, and the 

 bafe A B is equal to the given bafe. 



If the right line cut the circle in one point, it will alfo 

 cut it in two points, and therefore in this cafe there are two 

 triangles which anfwer the conditions ef the problem ; but 

 if it touches the circle only, then there is but one fuch 

 triangle ; and if the line F C E falls above the circle, then 

 the problem is impoffible. 



Prob. II. 



Having given the perimeter of a right-angled triangle* 

 and the perpendicular let fall from the right angle to the 

 oppofile fide, to conftruA it. Fig. 8. 



Anahfis. — Suppofe the thing done, and let A C B be the 

 propofed triangle ; produce the line .'^ B both ways to D 

 and E, making A D = A C, and C B = B E ; then will 

 D E reprefent the perimeter which is given by the quef- 

 tion ; join D E and C E. Then becaufe D A = A C, 

 the angle A D C = D C A ; but the angle C A B is equal 

 to the two angles ADC and D C A, or it is equal to 

 double the angle DC A: in the fame manner it may b<^ (hewn 

 that the angle A B C is equal to double the angle B C E ; 

 but the angles C A B and ABC are together equal to 

 a right angle, and confequentiy, fince thefe are double of 

 the angles D C A and B C E, it follows, that the fum of 

 the latter two angles is given, being equal to half a right 

 angle ; and therefore alfo the whole angle D C E is given, 

 being equal to a right angle and half a right angle ; there- 

 fore the locus of the point C is in the circumference of a 

 given circle. And fince the perpendicu'ar C G is alfo 

 given, the locus of the point C is the right line C F, pa- 

 rallel to the bafe A B, whence the point C is determined, 

 being found in the interfeAien of the right line C F and 

 the given fegment D C E. 



Vol. XXI. 



ConJlruSioH.-— On the right line D E, equal to the given 

 perimeter, defcribe a fegment capable of containing an angle 

 equal to a right angle and iialf a right angle ; and parallel 

 to D E, and at the given perpendicular dillance, draw the 

 right line F C cutting the fegment in C and C : join D C, 

 C E ; and from C draw alfo C A, A B, making the 

 angles DC A and BC E refpedtively eqial to the angles 

 C D A and C E B, fo (hall A C B be the triangle re- 

 quired. 



For fince the angle D CA is equal to the angle D A C, 

 the fide D A is equal to A C, and for the (air.e re^foa 

 the fide C B is equal to B E, and therefore the three 

 fides of the triangle ABC are equal to the whole D E, 

 or to the given perimeter ; alfo, fince the angle D C E is 

 equal to a right angle and half a right angle, the angles 

 C D A and C E B are together equal to half a right angle ; 

 but the angle C A B is double the angle C D A, and the 

 angle C B A is double the angle C E B, and confequentiy 

 thefe two together are equal to a right angle ; therefore 

 the third angle of the triangle A C B is a right angle. 

 Hence, fince the pei-pendicular C G is equal to the given 

 perpendicular, by conilruftion, and the fum of the thre« 

 fides A B, AC, B C, equal to the given perimeter, and 

 the angle AC B equal to the given angle, it follow* that 

 the triangle A B C is that which was to be conilrufted. 



This conftruftion ferves equally for any other triangle, 

 provided the vertical angle be given : and the limits of 

 poffibility are the fame as in the preceding problem. 



We will add one other example from Dr. Pembertoii's 

 paper on this fubjeiS, printed in vol. liii. of the Philofophical 

 Tranfaftions, and will then proceed to the conlideration of 

 loci of the higher orders. 



Prob. III. 



Let it be propofed to draw a triangle given in fpecies, fo 

 that two of its angles may touch a right line given in po- 

 fition, and the third angle a given point. 



This problem, which would be extremely difficult to folTC 

 algebraically, admits of more than one very concife geo- 

 metrical folution ; and as they will occupy but little fpace, 

 it is prefumed they will not be unacceptable to the reader 

 of this article. 



In the firll place, fuppofe a circle (fg. g.) to paf» 

 through the three points A, E, D, which Ihall interfeft A C 

 in G. Then E G, D G, being joined, the angle D EG 

 will be equal to the given angle D A C, both infilling on 

 the fame arc D G : alfo the angle E D G is the cornple- 

 ment to the two right ones of the given angle B A C : 

 thefe angles therefore are given, and the whole figure ' 

 E F G D given in fpecies. Confequentiy the angle E G F, 

 and its equal A D E will be given, together with the fide 

 D E of the triangle in pofition. 



Again, fuppofe a circle (fg. lO. ) to pafs through the 

 three points A, E, F, cutting A D in H, and E H, 

 F H joined. Here the angle E F H will be equal to the 

 given angle E A H ; and the angle F E H equal to the 

 given angle F A H. Therefore the whole figure E H F" D 

 15 given in fpecies ; and confequentiy the angle A D E at 

 before. 



Laftly, fuppofe a circle (fg. 11.) to circumfcribe the 

 triangle, and interfeft one of the lines, as A C in I. Then 

 D I being drawn, the angle D I F will be equal to the given 

 angle D E F in the triangle; confequentiy D I is inclined 

 to A C in a given angle, and is given in pofition, as alfo 

 the point I given ; whence I E being drawn, the angle 

 FIE will be the complement of the angle E D F in the 

 triangle to two right *bc9. Therefore I E is given in po- 



I i Ution, 



