Def.-A triangle which has no two sides equal to one another is a scalene triangle. Hence a scalene triangle has no two angles equal to one another. 58°. Theorem.-If two triangles have the three sides in the one respectively equal to the three sides in the other the triangles are congruent. B A A' A'C' along AC, and let B' fall at some point D. A'C'=AC, C' falls at C, B' and AADC is the AA'B'C' in its reversed position. Since AB=AD and CB=CD, (27°) A and C are on the right bisector of BD, and AC is the right bisector of BD. LBAC=✩DAC; and the As BAC and DAC are congruent. ▲ABC=▲A'B'C'. (54°) (53°, Cor. 2) (52°) q.e.d. 59°. Theorem.—If two triangles have two angles and the included side in the one equal respectively to two angles and the included side in the other, the triangles are congruent. LCLC the As ABC and A'B'C' are congruent. and A'C' AC) Proof.-Place A' on A, and A'C' along AC. C' LA'=LA, A'B' lies along AB; Because A'C' AC, C' coincides with C; (27°) and and (24°, Cor. 3) q.e.d. ¿C'=4C, C'B' lies along CB; and the triangles are congruent. A 60°. Theorem.-An external angle of a triangle is greater than an internal opposite angle. L B The external angle BCD is greater than the internal opposite angle ABC or BAC. Proof.-Let BF be a median produced until FG-BF. Therefore the LS BCD and ACE are each greater than each of the LS ABC and BAC. q.e.d. 61°. Theorem.--Only one perpendicular can be drawn to a line from a point not on the line. B Proof. Let B be the point and AD the line ; and let BC be 1 to AD, and BA be any line other than BC. BC is the only perpendicular from B to AD. q.e.d. Cor. Combining this result with that of 42° we haveThrough a given point only one perpendicular can be drawn to a given line. 62°. Theorem.--Of any two unequal sides of a triangle and the opposite angles— 1. The greater angle is opposite the longer side. 2. The longer side is opposite the greater angle. 2. LC is > LA; then AB is > BC. Proof. From the Rule of Identity (7°), since there is but one longer side and one greater angle, and since it is shown (1) that the greater angle is opposite the longer side, therefore the longer side is opposite the greater angle. q.e.d. Cor. 1. In any scalene triangle the sides being unequal to one another, the greatest angle is opposite the longest side, and the longest side is opposite the greatest angle. Also, the shortest side is opposite the smallest angle, and conversely. Hence if A, B, C denote the angles, and a, b, c the sides respectively opposite, the order of magnitude of A, B, C is the same as that of a, b, c. 63°. Theorem.—Of all the segments between a given point and a line not passing through the point— 1. The perpendicular to the line is the shortest. 2. Of any two segments the one which meets the line further from the perpendicular is the longer; and conversely, the longer meets the line further from the perpendicular than the shorter does. 3. Two, and only two segments can be equal, and they lie upon opposite sides of the perpendicular. P is any point and BC a line not passing through it, and PA is to BC. 1. PA which is to BC is shorter than any segment PB which is not L B to BC. 2. AC is > AB, then also PC is > PB. Proof. Since AC is > AB, let D be the point in AC so that AD=AB. Then A is the middle point of BD, and PA is the right bisector of BD. (42°, Def.) (53°) LPDB=LPBD. (53°, Cor 1) LPDB is > PCB; PBD is > PCB, PC is > PB. (62°, 2) q.e.d. The converse follows from the Rule of Identity. 3. Proof.—In 2 it is proved that PD=PB. Therefore two equal segments can be drawn from any point P to the line BC; and these lie upon opposite sides of PA. No other segment can be drawn equal to PD or PB. For it must lie upon the same side of the perpendicular, PA, as one of them. If it lies further from the perpendicular than this one it is longer, (2), and if it lies nearer the perpendicular it is shorter. Therefore it must coincide with one of them and is not a third line. q.e.d Def. The length of the perpendicular segment between any point and a line is the distance of the point from the line. 64°. Theorem.—If two triangles have two angles in the one respectively equal to two angles in the other, and a side opposite an equal angle in each equal, the triangles are congruent. If LA'=LA ] then the As B and A'B' ABJ are congruent. Proof-Place A' on A, and A'B' along AB. Also, . A DC B' A' (27°) (34°) Now if C' does not coincide with C, let it fall at some other point, D, on AC. Then, and But AB=A'B', AD=A'C', and LA=LA', .. ▲A'B'C'=^ABD, LADB=LC'. LC'=LC, (520) (52°, Cor.) (hyp.) LADB=LC, which is not true unless D coincides with C. Therefore C' must fall at C, and the As ABC and A'B'C' are congruent. The case in which D may be supposed to be a point on AC produced is not necessary. For we may then superimpose the ABC on the ▲A'B'C'. 65°. Theorem.-If two triangles have two sides in the one respectively equal to two sides in the other, and an angle opposite an equal side in each equal, then 1. If the equal angles be opposite the longer of the two sides in each, the triangles are congruent. 2. If the equal angles be opposite the shorter of the two C |