is equal to two right angles (B. I, Prop. xxiv, Cor. 2). But the sum of the two angles HDK and FDG is also equal to two right angles (B. I, Prop. 1); therefore the sum of the two angles HAK, HDK is equal to the sum of the two angles FDG, HDK; and taking the angle HDK from each of these equals, we have the angles HAK and FDG equal. In a similar manner it may be shown that the angles B and DFG are equal, also the angles C and DGF ; consequently these two triangles are similar. When one of the triangles is situated without the other, or partly without and partly within, we may construct a new triangle having its sides parallel with the sides of one of the given triangles, and lying wholly within the other; and since this new triangle is situated wholly within, it must be similar to the triangle in which it is situated. Consequently, by the first part of this proposition, the triangle which has its sides parallel to the sides of the new triangle must also be similar to the other given triangle. (66) Let it be required to find the width AB of a river, without crossing it. Take a position C in a direct line with A and B ; also take any point, as D, near the bank of the river. Then, having driven stakes at A, C and D, stretch a line from C to A, and thence to D: mark that point of the line which touches the stake at A. Remove the line, and exchange ends; that is, fasten to the stake D the end which was first made fast at C, and fasten upon C the end D,; then, taking hold of the line at the point which was in contact with A, stretch the line so that it shall take the position CFD. The figure ADFC is evidently a parallelogram, since, by construction, the opposite sides are equal. On the line AD, take the point G in a direct line with FB. The two triangles FDG and BAG are similar, since the angles ABG and DFG are alternate in reference to the parallels BC, DF, and are therefore equal (B. I, Prop. xvII); also the angle AGB is equal to FGD, being opposite angles (B. I, Prop. 11). These two triangles, having equal angles, are similar, and we have GD DF:: GA: AB. [B. IV, Prop. vi.] In this proportion, the first, second and third terms being known, we can find the fourth term, which is the distance sought. (67.) PROBLEM. To find the height of a perpendicular object, by means of an artificial horizon. An artificial horizon is a horizontal surface of any substance capable of reflecting light uniformly, as the surface of mercury, ink, etc. If soft treacle be placed in a saucer, it will form a, very good artificial horizon. B A the distances CF and FA; then, since the triangles FCD and FAB are obviously similar, we have CF CD: FA: AB. (68.) The perpendicular height of an object may be found by means of its shadow, by placing a stick, of known length, perpendicularly in the ground, and measuring the length of its shadow; then make the following proportion : Shadow of the stick shadow of the object height of the stick height of the object. THALES is said to have taught the Egyptians how to measure the height of the pyramids, by means of their shadows. (69.) PROBLEM. Suppose AB to be a tree, standing on the horizontal plane AC; it is required to find at what point it must break, so that, by falling, the top may strike the ground at C (See Art. 25). Let the height of the tree be denoted by h, and the distance AC by b; then BC will equal √i2 + h3, [B. II, Prop. vi.] and consequently CD BC § √b2 + h2. = Again, the two triangles BDF and CAB are similar, having the angle B common, and the angle BDF equal to the angle BAC, each being a right angle (B. IV, Prop. vII); therefore This point may be found arithmetically as follows: Draw DK parallel to AB, and FL perpendicular to AB. Let the height of the tree AC be denoted by h1, the height of the tree BD by ha, and the distance AB by d; then will KC = AC — BD = h1— h2, FL = ( h1 + ha). Since the sides of the two triangles DKC, FLG are respectively perpendicular to each other, they are similar (B. IV, Prop. ví), and we have DK: KC FL: LG; or, in symbols, In the two triangles ABC, DFG, if AB: DF: AC: DG :: BC: FG, the two triangles will have their corresponding angles equal. For, if the triangle ABC be not equiangular with the triangle DFG, suppose some other triangle having the same base DF, as DFH, to be equiangular with ABC. But this is impossible; for if the two triangles ABC, DFH were equiangular, their sides would be proportional (B. IV, Prop. vII), and we should have AB: DF: AC: DH; but, by supposiAB: DF :: AC: DG, and therefore DH = DG. We should also have tion, tion, FG. AB: DF BC: FH; but, by supposiAB: DF: BC: FG, and therefore FH = Hence the two triangles DFH and DFG are identical, since all the sides of the one are equal to all the sides of the other (B. I, Prop. vIII); which is absurd, since their angles are unequal. Therefore the triangles ABC and DFG must be equiangular. (71.) THEOREM. The sum of the squares of the diagonals of a quadrilateral is twice the sum of the squares of the two lines bisecting the opposite sides. B F H G divided proportionally, each being halved, the line FG must be parallel to AC (B. IV, Prop. п). In the same way we may show that HK is parallel to AC, and therefore FG and HK are parallel. Furthermore we shall have GH and FK each parallel to BD, and therefore parallel to each other. Consequently the figure FGHK is a parallelogram (Def. XVII). Again, since the sides of the triangles FAK, BAD are respectively parallel, they are similar, and BA : FA :: BD: FK; and since BA is twice FA, it follows that BD is twice FK. In the same way, it may be shown that AC is twice FG. Hence BD24 FK2 2 FK2 + 2 GH2; AC2 = 4 FG22 FG2 + 2 HK2. Consequently AC2 + BD2 = 2 (FG2 + GH2 + HK2 + KF2) 2 (FH+GK'). [B. II, Prop. XIV.] Cor. 1. From the above demonstration, it follows, that if the points. of bisection of the four sides of a trapezium be joined, we shall thus form a parallelogram. |