They in all their parts (P. V.). But this is impossible, because a part cannot be equal to the whole (A. 8): hence, the hypothesis that AB and AC are unequal, is false. must, therefore, be equal; which was to be proved. Cor. An equiangular triangle is equilateral. PROPOSITION XIII. THEOREM. In any triangle, the greater side is opposite the greater angle; and, conversely, the greater angle is opposite the greater side. In the triangle ABC, let the angle ACB be greater than the angle ABC: then will the side AB be greater than the side AC. For, draw CD, making the angle BCD equal to the angle B (Post. 7): then, in the triangle DCB, we have the angles DCB and DBC equal: hence, the opposite sides DB and DC are equal (P. XII.). In the triangle ACD, we have (P. VII.), or, since DC = DB, and AD + DB = AB, we have, which was to be proved. AB > AC; Conversely: Let AB be greater than AC: then will the angle ACB be greater than the angle ABC. For, if ACB were less than ABC, the side AB would be less than the side AC, from what has just been proved; if ACB were equal to ABC, the side AB would be equal to AC, by Prop. XII.; but both conclusions are contrary to the hypothesis: hence, ACB can neither be less than, nor equal to, ABC; it must, therefore, be greater; which was to be proved. PROPOSITION XIV. THEOREM. From a given point only one perpendicular can be drawn to a given straight line. Let A be a given point, and AB a perpendicular to DE: then can no other perpendicular to DE be drawn from A. F For, suppose a second perpendicular AC to be drawn. Prolong AB till BF is equal to AB, and draw CF. Then, the triangles ABC and FBC will have AB equal to BF, by construction, CB common, and the included angles ABC and FBC equal, because both are right angles hence, the angles ACB and FCB are equal (P. V.) But ACB is, by a hypothesis, a right angle hence, FCB must also be a right angle, and consequently, the line ACF must be a straight line (P. IV.). But this is impossible (A. 11). The hypothesis that two perpendiculars can be drawn is, therefore, absurd; consequently, only one such perpendicular can be drawn; which was to be proved. В B. C If the given point is on the given line, the proposition is equally true. For, if from A two perpendiculars AB and AC could be drawn to DE, we should have BAE and CAE each equal to a right angle; and consequently, equal to each other; which is absurd (A. 8). D -E PROPOSITION XV. THEOREM. If from a point without a straight line a perpendicular be let fall on the line, and oblique lines be drawn to different points of it: 1o. The perpendicular will be shorter than any oblique line. 2o. Any two oblique lines that meet the given line at points equally distant from the foot of the perpendicular, will be equal: 3°. Of two oblique lines that meet the given line at points unequally distant from the foot of the perpendicular, the one which meets it at the greater distance will be the longer. Let A be a given point, DE a given straight line, AB a perpendicular to DE, and AD, AC, AE oblique lines, BC being equal to BE, and BD greater than BC. Then will AB be less than any of the oblique lines, AC will be equal to AE, and AD greater than AC. Prolong AB until BF is equal to AB, and draw FC, FD. 1o. In the triangles ABC, FBC, we have the side AB equal to BF, by construction, the side BC common, and the included angles ABC and FBC equal, because both are right angles: hence, FC is equal to AC. (P. V.). But, AF is shorter than ACF (A. 12): hence, AB, the half of AF, is shorter than AC, the half of ACF; which was to be proved. 2o. In the triangles ABC and ABE, we have the side BC equal to BE, by hypothesis, the side AB com mon, and the included angles ABC and ABE equal, because both are right angles: hence, AC is equal to AE; which was to be proved. 3°. It may be shown, as in the first case, that AD is equal to DF. Then, because the point C lies within the triangle ADF, the sum of the lines AD and DF will be greater than the sum of the lines AC and CF (P. VIII.): hence, AD, the half of ADF, is greater than AC, the half of ACF; which was to be proved. Cor. 1. The perpendicular is the shortest distance from a point to a line. Cor. 2. From a given point to a given straight line, only two equal straight lines can be drawn; for, if there could be more, there would be at least two equal oblique lines on the same side of the perpendicular; which is impossible. If a perpendicular be drawn to a given straight line at its middle point: 1°. Any point of the perpendicular will be equally distant from the extremities of the line: 2o. Any point, without the perpendicular, will be unequally distant from the extremities. Let AB be a given straight line, C its middle point, and EF the perpendicular. Then will any point of EF be equally distant from A and B; and any point without EF, will be unequally distant from A and B. A 1o. From any point of EF, as D, draw the lines DA and DB. Then will DA and DB be equal (P. XV.): hence, D is D E B equally distant from A and B ; which was to be proved. 2o. From any point without EF, as I, draw IA and IB. One of these lines, as IA, will cut EF in some point D; draw DB. Then, from what has just been shown, DA and DB will be equal; but IB is less than the sum of ID and DB (P. VII.); and because the sum of ID and DB is equal to the sum of ID and DA, or IA, we have IB less than IA: hence, I is unequally distant from A and B; which was to be proved. A B Cor. If a straight line EF have two of its points E and F equally distant from A and B, it will be perpendicular to the line AB at its middle point. PROPOSITION XVII. THEOREM. If two right-angled triangles have the hypothenuse and a side of the one equal to the hypothenuse and a side of the other, each to each, the triangles will be equal in all their parts. A Let the right-angled triangles ABC and DEF have the hypothenuse AC equal to DF, and the side AB equal to DE: then will the triangles be equal in all their parts. B D the triangles will be Let us suppose then, If the side BC is equal to EF, equal, in accordance with Proposition X. that BC and EF are unequal, and that BC is the longer. On BC lay off BG equal to EF, and draw AG. The triangles ABG and DEF have AB equal to DE by hypothesis, BG equal to EF, by construction, and |