14 



♦ KNOVS^LEDGE 



[JcLY 6, 1883. 



&UV iHattjtmatiral Column* 



GEOMETRICAL PKOBLEMS. 



By Richard A. Proctor. 



Part VI. 



LET ns next try a few problems — properly so termed — that is, 

 propositions in whicli something is required to be done. In 

 these, as we have said, the analytical method is nearly always to 

 be preferred. We will begin with a simple example. 



Ex. 6. — On a given straight line describe an isosceles triangle, each 

 of whose equal sides shall he double of the base. 



Fig. 11. 



Let A B (Fig. 11) be the given straight line. 



Suppose that what is required is done, and that on the base, A B, 

 there has been described the triangle, A C B, in which the sides 

 A C and C B are equal to each other, and each double of the base 

 A B ; and let us consider what constructions are suggested. 



It seems hardly possible that the resemblance between this 

 problem and Euc. I., 1, should escape the student's notice. He 

 will inquire, then, whether the method of that problem cannot be 

 applied to the present cue. Instead of the circle with radius 

 equal to A B, we now require circles with radius equal to twice 

 A B. It is clear, then, that if we produce A B to D, making B D 

 equal to A B, and B A to E, making A E equal to A B (Euc. L, 3), 

 then A D and B E will each be double of A B.* Therefore, if with 

 centre A and radius A D we describe a circle D C F, and with 

 centre B and radius B E the circle E C G, then C, the intersection 

 of these circles, is the vertex of the required triangle. For A C 

 and B C are severally equal to AD and E B— that is, are double of 

 the base, A B. 



We will next trj' the following : — 



Ex. 7. — The point P, Fig. 12, is within the acute angle formed by 

 the lines A B and A C. It is required to draiv through P a straight 

 line which shall cut off equal parts from A B and A C. 



Fig. 12. 



Let D P E be the required line, so that A D is equal to A E.f 



Then D A E is an isosceles triangle, and it is an obvious cotrrse to 



see whether any of the properties of isosceles triangles will help us 



to a solution of our problem. Now, the only property of isosceles 



triangles explicitly contained in Euclid is that of Bk. I., prop. 5. 



* We have seen this problem given with the proviso that no 

 problem beyond Euc. I., 1, shall be made use of. In this case the 

 student will see at once that if, ■(vith centres A and B, and distance 

 A B, he describes the circles B F E, A G D, then E B and A D, the 

 diameters of these equal circles, are severally double of A B. 



t In constructing the figure, proceed thus : — Take A D equal to 

 A E, and join D E ; then take P, a point dividing D E into unequal 

 parts. 



This gives us the angle A D E equal to the angle AE D, — a property 

 which avails us nothing. 



But there are other properties of isosceles triangles, not expressly 

 mentioned by Euclid, which every geometrician ought to be ac- 

 quainted with. We will assume that the student is familiar with 

 them — and indeed they are nearly self-evident. They are included 

 in the statement that the perpendicular from the vertex on the 

 base of an isosceles triangle bisects the base and also the vertical 

 angle. Draw A il perpendicular to the assitmed line D E ; then 

 the angle M A D is equal to the angle MAE, and also D M is 

 equal to M E. 



Now let us consider whether this construction affords us any 

 hints : — 



First, we cannot see how to di-aw the line through A perpendi- 

 cular to the real line D E, because it is this very line we seek to 

 draw. 



Secondly, we cannot, for n similar reason, see how to draw the 

 line from A to the bisection ol li K. 



But, thirdly, we can draw the line A M, bi-secting the angle 

 DAE. 



And this clearly gives us the solution of otu- problem, since we 

 can now draw D P E at right angles to A 31. Thus the solution 

 runs as follows : — 



Draw A M bisecting the angle DAE, and through P draw 

 D P JI E at right angles to A M ; then shall A D be equal to A E. 

 For, in the triangles M A D, 31 A E, the angle 31 A D is equal to the 

 angle MAE, the right angle A 31 D is equal to the right angle 

 A M E, and A M is common to the two triangles ; therefore the 

 triangles are equal in all respects (Euc. I., 26), and A D is equal 

 to AE. 



The proof of the equality of the triangles 31 A D and MAE 

 was not included in the prior examination of the problem, since 

 is involved in the assumed knowledge on the student's part of the 

 fundamental properties of isosceles triangles, proved farther on. 

 But, of course, it is well (in a case of such simplicity) to introduce 

 the proof into the solution of the problem. 



Let us next try the following problem : — 



Ex. 8. — The points P and Q, Fig. 13, are on the same side of the 

 line, AB. It is required to determine a point C in AB, such that 

 the lines P C QC may malce eqv/il angles with A B. 



Let C be the required point, so that the angle P C A is equal to 

 the angle Q C B.* 



Let us try drawing a Une, C D, at right angles to A B. Then the 

 angle P C D is equal to the angle Q C D. On a consideration of this 

 relation, however, it seems unlikely to help us. For it is not 

 easier to gather anything from the equality of P C D and Q C D, 

 than to make use of the equality of P G A and Q C B. 



It seems an obvious resource, since the equality of the angles, 

 P C A and Q C B, as they stand, is not readily applicable to our 

 purposes, to produce either P C oi Q C, in order to see whether the 

 vertical angle either of P C A or Q C B might be more serviceable 

 to us. Produce P C to E. Then the angles Q C B and B C E are 

 equal, or C B is the bisector of the angle Q C E. The only property 

 connected with the bisector of an angle which seems likely to help 

 us is this one, that the bisector of the vertical angle of an isosceles 

 triangle is perpendicular to and bisects the base. Now, we can 

 make an isosceles triangle of which C shall be the vertex and C Q a 

 side, for we have only to take C E equal to C Q, and to join Q E, 

 cutting C B in F. Then, by the property just mentioned, Q E is at 

 right angles to C F, and is bisected in F. 



These relations obviously supply all we want. For, reversing our 

 processes, we have only to draw Q F E perpendiculsir to A B, and 



* Construct as follows : Draw A B, and from any point C in A B, 

 (\Taw tbe uneqiial lines C P, C Q equally inclined to AB. Then 

 there is no risk that accidental relations will appear as necessary 

 ones. 



