288 



THE POPULAR EDUCATOR. 



- 30. 



which the learner is shown the method of drawing any triangle 

 having its sides equal to three given straight lines; but the 

 second, in which the length of the two equal sides and the alti- 

 tude of the triangle are the data given, requires further explana- 

 tion, and brings us to 



PROBLEM XXI. To draw an isosceles triangle of which the 

 length of the two equal sides and the altitude are given. 



Let A represent the length of the two equal sides, and B the 

 altitude of the isosceles triangle required. First draw the line 

 c D of indefinite length, and through the point E, taken as nearly 

 as possible in the centre of the line as drawn, draw the straight 

 line F G perpendicular, or at right angles to c D. From the 



point E along the straight 

 line E G set off a straight line 

 E H equal to B, and from the 

 point H along the straight line 

 H F set off H K equal to A. 

 Then from the point H, as 

 centre, with the distance H K, 

 describe the arc L K M, cut- 

 ting the straight line c D in 

 the points L, M. Join H L, 

 H M. The triangle H L M is 

 the isosceles triangle required, 

 for the length of its altitude, 

 H E, is equal to B, and the 

 length of its equal sides H L, 

 H M, is equal to A. 



PROBLEM XXII. To draw 

 an isosceles triangle of ivhich 

 the length of the base and the 

 altitude are given. 



In the above figure (Fig. 30), 

 i e t B) as b e f or e, represent the 

 altitude of the isosceles triangle required, and x the length of 

 its base. First draw the lino c D of indefinite length, and 

 within its limits set off a straight line L M equal to x. Bisect 

 L M in E, and at the point E draw E G perpendicular or at right 

 angles to c D, and from the point E, along the straight line E G, 

 set off E H equal to B. Join H L, H M. The triangle H L M is 

 the isosceles triangle required, for it has its base L M equal to x, 

 while its altitude, E H, is equal to B. 



By the aid of Fig. 30 we may easily discover some more facts 

 in geometry, which the student may prove to be correct to his 

 satisfaction by means of his compasses and parallel ruler. 



First join L K, and bisect L K in the point N. Join H N. The 

 straight line H N bisects the angle L H K, or divides it into the 

 two equal angles L H N, N H K. Now apply the parallel ruler 

 to the straight line H N, and by its aid draw through the point L 

 a straight line L o parallel to H N. This straight line L o meets 

 the straight line E G in the point o, and if the circumference of 

 the circle of which the arc L K M is a part, be completed, it will 

 also pass through the point o, in which the straight line L o 

 meets the straight line E G. Now by Theorem 2 (page 156) 

 when a straight line intersects two parallel straight lines the 

 alternate angles are equal, therefore the alternate angles NHL, 

 H L o, formed by the intersection of the straight line H L 

 with the parallel straight lines H N, o L, are equal to one 

 another. But since the triangle L H o is an isosceles triangle, 

 of which the side H o is equal to the side H L, being radii of the 

 same circle, the angle H L o is equal to the angle L o H or L o K 

 (as it does not matter whether we call the opening between the 

 lines o L, o K, the angle L o K or L o H), and as the angle I. H N 

 was shown to be equal to the angle H L o, it must be also equal 

 to the angle L o K. Now the angle L H K is double of the angle 

 L H N. Therefore the angle L H K is also double of the angle 



L O E. 



The next thing to be observed is that the angles L H K, L o K, 

 each stand on the same base L K, and that one of them, the 

 angle L H K, has its apex or vertex H at the centre H of the 

 circle o L K M, while the other, the angle L o K, has its vertex or 

 apex o on the circumference of the circle o L K M. And the 

 geometrical fact to be deduced from this is, that when two 

 angles stand on the same base, and on the same side of it, one 

 having its vertex at the centre of a circle and the other having 

 its vertex at the circumference of the same circle, the angle 

 which has its vertex at the centre is double of that which has 

 its vertex at the circumference. This is true at whatever point 



of the circumference the vertex of the angle at the circumference 

 may be, the term circumference being understood to apply to 

 that part of the whole circumference of the circle which lies on 

 the same side of the base as that on which the angles are found, 

 as the arc L o M of the whole circumference of the circle o L K M. 

 Thus the angle L H M, standing on the base L M, and having its 

 vertex at the centre H of the circle o L K M, is double of the angle 

 L o M, which stands on the same base and has its angle at the 

 circumference. It is also double of the angles L p M, L Q M, which, 

 have their vertices at the points P, Q, of the arc L o M. The angles 

 L P M, L 6 M, L Q M, being each of them equal to half of the angle 

 L H M, are equal to one another, from which we learn another 

 geometrical fact, namely, that all angles standing on the same 

 base and on tlie same side of it, and having their tops or vertices 

 in the circumference of a circle, are equal to one another. 



In Case 4, where the angle at the vertex of the triangle is 

 given, and the length of the two equal sides, all that is necessary 

 to be done is to draw an angle of the opening required bj 

 Problem VII. (page 191), and to set off the length of the two 

 equal sides along the legs of the angle, joining the points in 

 which the legs of the angle are cut in order to form the base ; 

 and in Case 10, when the angle of the vertex of the triangle is 

 given, but the length of the equal sides is not stated, the 

 triangle may be completed by cutting the legs of the angle in 

 any two points equidistant from the apex, and joining these 

 points to form the base as before. Case 5, however, on which 

 the length of the base and the angle at the apex of the triangle 

 is given, will require explanation in 



PROBLEM XXIII. To draw an isosceles triangle of which the 

 angle at the vertex of the triangle and the length of tlic base are 

 given. 



Let A be the angle at the vertex of the isosceles triangle 

 required, and let B represent its base. Draw any straight line, 

 c E, of indefinite length, and along c E set off c D equal to B. 

 Then at the point D in the straight line E D make the rectilineal 

 angle E D F equal to the given angle A by Problem VII. 

 (page 191) ; bisect c D in G, and through G draw G H perpen- 

 dicular to c D or c E. Now, because the three interior angles of 

 a triangle are equal to two right angles, the three interior angles 

 of the isosceles triangle required are together equal to the two 

 angles c D F, F D E, of which F D E is equal to the angle at the 

 vertex ; and as the angles at the base of an isosceles triangle 

 are equal, each of the remaining angles is equal to half of the 

 angle CDF. Bisect the angle c D F, by Problem VI. (page 191), 

 by the line D K, and from the 

 point K in which the straight 

 line D K cuts the perpendi- 

 cular G H, draw the straight 

 line K c to the extremity c 

 of the base c D. The triangle 

 K c D is the isosceles triangle 

 required, for its base c D is 

 equal in length to B, and the 

 angle c K D at the vertex of 

 the triangle is manifestly 

 equal to the given angle A. 



For Case 6, when the 

 angle at the vertex of the 

 triangle and the altitude 

 are given, if in Fig. 31 

 the straight line x repre- 



Fig. 31. 



sents the altitude, it is manifestly only necessary to make the 

 angle c K D equal to the given angle A, and then bisect it by 

 the straight line K L, and after setting off K G along the straight 

 line K L equal to the given altitude x, to draw c D through the 

 point G at right angles to K G, cutting the legs K c, K D, of the 

 angle c K D in the points c and D. The triangle K c D is of the 

 required altitude, and has the angle c K D at its vertex equal to 

 the given angle A. 



From what has been already said in Problems XXI., XXII., 

 and XXIII., the student will find no difficulty in forming 

 isosceles triangles under the conditions or data set forth in 

 Cases 7, 8, 9, and 11, which will afford useful exercises for 

 practice. The mode of construction is in all cases the same 

 whether the isosceles triangle be a right-angled triangle, an 

 obtuse-angled triangle, or an acute-angled triangle ; or in other 

 words, whether it have a right angle, an obtuse angle, or an 

 acute angle at its vertex. 



