For if it were less, then the opposite side would be less (293), and if it were equal, then the opposite sides would be equal (284); both of which are contrary to the hypothesis. PROBLEMS IN DRAWING. 295. Problem.-To draw a triangle when the three sides are given. Let a, b, and c be the given lines. Draw the line IE equal to c. With I as a center, and with the line b as a radius, describe an arc, and with E as a center and the line a as a radius, describe a second arc, so that the two may cut each other. Join O, the point of intersection of these arcs, with I and with E. IOE is the required triangle. a b a E If c were greater than the sum of a and b, what would have been the result? What, if c were less than the difference of a and b? Has the problem more than one solution; that is, can unequal triangles be drawn which comply with the conditions? Why? 296. Corollary. In the same way, draw a triangle equal to a given triangle. 297. Problem.-To draw a triangle, two sides and the included angle being given. Let a and b be the given lines, and E the angle. Draw FC equal to b. an angle equal to E. equal to a, and join FD. a At C make Take DC Then FDC Is this problem always soluble, whatever may be the size of the given angle, or the length of the given lines? Why? 298. Problem.-To draw a triangle when one side and the adjacent angles are given. Let a be the given line, and D and E the angles. Draw BC equal to a. At B make an angle equal to D, and at C an angle equal to E. Produce the sides till they meet at the point F. FBC is a triangle having the given side. and angles. Has this problem more than one solution? Can it be solved, whatever be the given angles, or the given line? B a 299. Problem.-To draw a triangle when one side and two angles are given. If one of the angles is opposite the given side, find the supplement of the sum of the given angles (214). This will be the other adjacent angle (256). Then proceed as in Article 298. 300. Problem.-To draw a triangle when two sides and an angle opposite to one of them are given. Construct an angle equal to the given angle. Lay off on one side of the angle the length of the given adjacent side. With the extremity of this adjacent side as a center, and with a radius equal to the side opposite the given angle, draw an arc. This arc may cut the opposite side of the angle. Join the point of intersection with the end of the adjacent side which was taken as a center. A triangle thus formed has the required conditions. The student can better discuss this problem after drawing several triangles with various given parts. Let the given angle vary from very obtuse to very acute; and let the opposite side vary from being much larger to much smaller than the side adjacent to the given angle. Then let the student explain when this problem has only one solution, when it has two, and when it can not be solved. EXERCISES. 301.-1. The base of an isosceles triangle is to one of the other sides as three to two. Find by construction and measurement, whether the vertical angle is acute or obtuse. 2. Two right angled triangles are equal, when any two sides of the one are equal to the corresponding sides of the other. 3. Two right angled triangles are equal, when an acute angle and any side of the one are equal to the corresponding parts of the other. 4. Divide a given triangle into four equal parts. 5. Construct a right angled triangle when, I. An acute angle and the adjacent leg are given; SIMILAR TRIANGLES. 302. Similar magnitudes have been defined to bethose which have the same form while they differ in extent (37). 303. Let the student bear in mind that the form of a figure depends upon the relative directions of its points, and that angles are differences in direction. Therefore, the definition may be stated thus: Two figures are similar when every angle that can be formed by lines joining points of one, has its corresponding equal and similarly situated angle in the other. ANGLES EQUAL. 304. Theorem.-Two triangles are similar, when the three angles of the one are respectively equal to the three angles of the other. This may appear to be only a case of the definition of similar figures; but it may be shown that every angle that can be made by any lines whatever in the one, may have its corresponding equal and similarly situated angle in the other. Let the angles A, B, and C be respectively equal to the angles D, E, and F. Suppose GH and IR to be any two lines in the triangle ABC. Join IC and GR. From F, the point homologous to C, extend FL, making the angle LFE equal to ICB. Now, the triangles LFE and ICB have the angles B and E equal, by hypothesis, and the angles at C and F equal, by construction. Therefore, the third angles, ELF and BIC, are equal (262). By subtraction, the angles AIC and DLF are equal, and the angles ACI and DFL. From L extend LM, making the angle FLM equal to CIR. Then the two triangles FLM and CIR have the angles at C and F equal, as just proved, and the angles at I and L equal, by construction. Therefore, the third angles, LMF and IRC, are equal. Join RG. Construct MN homologous to RG, and NO homologous to GH. Then show, by reasoning in the same manner, that the angles at N are equal to the corresponding angles at G; and so on, throughout the two figures. The demonstration is similar, whatever lines be first made in one of the triangles. Therefore, the relative directions of all their points are the same in both triangles; that is, they have the same form. Therefore, they are similar figures. 305. Corollary.-Two similar triangles may be divided into the same number of triangles respectively similar, and similarly arranged. 306. Corollary.-Two triangles are similar, when two angles of the one are respectively equal to two angles of the other. For the third angles must be equal also (262). 307. Corollary.-If two sides of a triangle be cut by a line parallel to the third side, the triangle cut off is similar to the original triangle (124). 308. Theorem.-Two triangles are similar, when the sides of one are parallel to those of the other; or, when the sides of one are perpendicular to those of the other. We know (138 and 139) that the angles formed by lines which are parallel are either equal or supplementary; and that the same is true of angles whose sides are perpendicular (140). We will show that the angles can not be supplementary in two triangles. If even two angles of one triangle could be respect ively supplementary to two angles of another, the sum of these four angles would be four right angles; and then the sum of all the angles of the two triangles would be more than four right angles, which is impossible (255). Hence, when two triangles have their sides respectively parallel or perpendicular, at least two of the angles of one triangle must be equal to two of the other. Therefore, the triangles are similar (306). |