16. Construct a scalene triangle ABC, and on the side of BC remote from A, describe a triangle DBC with DB= AC, and DC = AB. Join AD. How do AD and BC appear to divide each other? Repeat the construction several times with different pairs of triangles, and note whether the same peculiarity of division occurs in each case. 17. Construct a scalene triangle ABC. On the other side of BC construct DBC with DB= AB, and DC=AC; on the other side of AC construct EAC with CE=CB, and AE=AB; on the other side of AB construct FAB with BF-BC, and AF-AC. Join AD, BE, CF. What lines in the figure are bisected? What triangles are isosceles? What angles are right angles? How many right-angled triangles are there? 18. Construct a triangle ABC (BC=47, CA=40, AB=27 millimetres). On the other side of BC construct DBC with DB= AC, and DC=AB; on the other side of AC construct EAC with EC AB, and EA = BC; on the other side of AB construct FAB with FA =BC, and FB AC. Join AD, BE, CF. What are the positions of DC, and CE with respect to each other; also EA, AF; and FB, BD? What lines in the figure are bisected? Has any line the third part cut off? Has any line the sixth? 19. If two sides of a triangle are unequal, the angles opposite to them are unequal. (Suppose the angles equal and prove an absurdity.) 20. If two angles of a triangle are unequal, the sides opposite to them are unequal. CHAPTER III. Equality of Triangles. 1. Construct two triangles, each with sides of lengths 11, 11 and 12 inches, as indicated in the adjacent figures. Adjust the bevel, or the protractor, to the angle A, and also to the angle D, and carefully compare the magnitudes of these angles. In like manner compare the magnitudes of the angles B and E, and also the magnitudes of the angles C and F. Next cut both triangles from the paper, and place one triangle upon the other so that the corresponding angular points coincide. From this superposition what conclusion do you draw as to the areas of the triangles? Repeat the same construction, measurement, and superposition with two triangles whose sides are 4, 2 and 4 inches; with two whose sides are 50, 80 and 100 millimetres; etc. The result of our observations in these cases is that if two triangles have their sides equal, the angles which are opposite to equal sides are equal, and the areas are equal. In other words two such triangles are the same triangle in different positions. Another way of stating the fact is to say that if the sides of a triangle are fixed, the angles are fixed, and the area is fixed. 2. Construct two angles, BAC and EDF, each of 30°. On sides of these angles measure off distances AB and DE, each of length 40 millimetres; and also distances AC and DF, each of length 51 millimetres. Join BC and EF, thus forming two triangles, ABC and DEF. Adjust the bevel, or protractor, to the angle B, and also to the angle E, and carefully compare the magnitudes of these two angles. In like manner compare the magnitudes of the angles C and F. With the dividers compare the magnitudes of the sides BC and EF. Further, cut one triangle from the paper, and place it upon the other. From this superposition what conclusion do you draw as to the areas of the two triangles? Repeat the same construction, measurement, and superposition with the following triangles: Two whose sides are 13 and 24 inches, and included angle 30°. Two whose sides are 30 and 110 millimetres, and included angle 78°. Two whose sides are 1 and 2 inches, and included angle 135°. The result of our observations in all these cases is that if two triangles have two sides in each equal, and the angles included by these two sides equal, then the remaining sides are equal, and the angles opposite to equal sides are equal, and the triangles are equal in area. In other words two such triangles are the same triangle in different positions. Another way of stating the fact is to say that if two sides and the included angle of a triangle are fixed, the remaining side and angles are fixed, and the area is fixed. к 3. In the case of all the triangles in 1 and 2, lay off, with the bevel, three angles adjacent to one another, equal to the three angles of each triangle, in the way indicated in the figure. Determine the positions of the initial and final lines, LM and LK, with respect to one another. K M The result of such an examination will be found to be that the lines KL and LM are in the same straight line, i.e., the sum of the three angles in each of these triangles is two right angles, or 180°. 4. It is proposed to show that the sum of the three angles of any triangle must be two right angles, or 180° : the position FG, and turn it through the angle CAB, to the position HK. Slide it along AB to the position BL, and turn it through the angle B, to the position BM. The pencil has rotated through all the angles of the triangle. But in its final position BM it points in a direction just opposite to its first position DC, and therefore must have rotated through 180°. Hence all the angles of this (which is any) triangle must together equal 180°, or two right angles. It follows that if two triangles have two angles in the one equal to two angles in the other, the third angle in one triangle is equal to the third angle in the other. |