The equal angles BÂC, EDF are opposite the corresponding sides BC, EF. The equal angles ABC, DÊF are opposite the corresponding sides CA, FD. The equal angles BĈA, EFD are opposite the corresponding sides AB, DE. Art. 106. NOTE ON PROPOSITION 27. The proviso that the sides of the triangles are proportional when taken in order is very important. It is quite possible for the sides of one triangle to be proportional to the sides of another without the triangles being similar. Suppose that in the triangles ABC, DEF, then it may be proved (see Proposition 56 below) that CA: AB=FD : FE. But the triangles are not similar. In the first proportion BC corresponds to EF. In the second proportion BC corresponds to FD. Hence the sides of the two triangles cannot be made to correspond. Art. 107. PROPOSITION XXVIII. (Euc. VI. 6.) ENUNCIATION. If two sides of one triangle be proportional to two sides of another triangle, and if the included angles be equal, then the triangles are similar; and those angles are equal which are opposite to corresponding sides. and In the triangles ABC, DEF let it be given that BA : AC = ED : DF, it is required to prove that the triangles are similar; and that the angles BCA, EFD opposite the corresponding sides BA, ED are equal; and that the angles ABC, DEF opposite the corresponding sides AC, DF are equal. From A, the vertex of the triangle ABC which corresponds to D, measure off on AB, the side corresponding to DE, a length AG equal to DE. Draw GH parallel to BC cutting AC at H. It will first be proved that the triangles AGH, DEF are congruent. Hence the angles of the triangle DEF are respectively equal to the angles of the triangle ABC. Hence by Prop. 26 the triangles DEF, ABC are similar, and those angles are equal which are opposite to corresponding sides. Art. 108. EXAMPLES. 34. Two parallel straight lines are cut by any number of straight lines passing through a fixed point. Prove that the intercepts made on the parallel lines by any two of the straight lines through the fixed point have a constant ratio. 35. If the tangents at A and B to a circle meet at C, and if P be any point on the circle, and if PQ, PR, PS be drawn perpendicular to AC, CB, BA respectively, then prove that the triangles PAS, PBR are similar; and that the triangles PBS, PAQ are similar; and that PS is a mean proportional between PQ and PR. 36.* If C be the centre of a circle, F any point outside it, if FA, FB be tangents to the circle at A and B respectively, if FP be any straight line through F cutting the circle at P; and if through P a straight line be drawn perpendicular to FP cutting CA at Q and CB at R; then prove that the triangles CFQ, CRF are similar; and that CF is a mean proportional between CQ and CR. 37. Let O be the centre of a circle, and C a fixed point in its plane. Let CO cut the circle at A and B. Let P be any point on the circle, and through P let a straight line be drawn perpendicular to CP, cutting the tangents at A and B at Q and R respectively, then prove that (1) the triangles ACQ, BCR are similar. (2) AQ: AC BC: BR. (3) the angle QCR is a right angle. Art. 109. PROPOSITION XXIX. (Euc. VI. 7.) ENUNCIATION. If two triangles have one angle of the one equal to one angle of the other, and the sides about one other angle in each proportional in such a manner that the sides opposite to the equal angles correspond, then the triangles have their remaining angles either equal or supplementary, and in the former case the triangles are similar. On AB, the side corresponding to DE, take a length AG equal to DE, and draw GH parallel to BC cutting AC at H. The triangles AGH, DEF will first be compared. *Notice that the sides about the angles BAC, EDF are proportional in such a manner that the sides AC, DF opposite the equal angles ABC, DEF correspond. Notice further that in each triangle two angles have been referred to, viz. ABC, BẤC in the triangle ABC and DEF, EDF in the triangle DEF, and therefore the remaining angles are ACB, DÊE. |