For the straight line AB is the shortest path between the points A and B (Def. 3); hence AB is less than the sum of AC and BC. For the same reason, BC is less than the sum of AB and AC; and AC less than the sum of AB and BC Therefore, any two sides, &c. PROPOSITION IX. THEOREM. If, from a point within a triangle, two straight lines are drawn to the extremities of either side, their sum will be less han the sum of the other two sides of the triangle. Let the two straight lines BD, CD be drawn from D, a point within the triangle. ABC, to the extremities of the side BC; then will the sum of BD and DC be less than the sum of BA, AC, the other two sides of the triangle. Produce BD until it meets the side AC B A C in E; and, because one side of a triangle is less than the sum of the other two (Prop. VIII.), the side CD of the triangle CDE is less than the sum of CE and ED. To each of these add DB; then will the sum of CD and BD be less than the sum of CE and EB. Again, because the side BE of the triangle BAE is less than the sum of BA and AE, if EC be added to each, the sum of BE and EC will be less than the sum of BA and AC. But it has been proved that the sum of BD and DC is less than the sum of BE and EC; much more, then, is the sum of BD and DC less than the sum of BA and AC, Therefore, if from a point, &c. PROPOSITION X. THEOREM. The angles at the base of an isosceles triangle are equal to one another. Let ABC be an isosceles triangle, of which the side AB is equal to AC; then will the angle B be equal to the angle C. D C For, conceive the angle BAC to be bisected by the straight line AD; then, in the two triangles ABD, ACD, two sides AB, AD, and the included angle in the one, are equal to the two B sides AC, AD, and the included angle in the other; there fore (Prop. VI.), the angle B is equal to the angle C. The efore, the angles at the base, &c. * Cor. 1. Hence, also, the line BD is equal to DC, and the angle ADB equal to ADC; consequently, each of these angles is a right angle (Def. 10). Therefore, the line bisecting the vertical angle of an isosceles triangle bisects the base at right angles; and, conversely, the line bisecting the base of an isosceles triangle at right angles bisects also the vertical angle, Cor. 2. Every equilateral triangle is also equiangular. Scholium. Any side of a triangle may be considered as its base, and the opposite angle as its vertex; but in an isos celes triangle, that side is usually regarded as the base, which is not equal to either of the others. PROPOSITION XI. THEOREM (Converse of Prop. X.). If two angles of a triangle are equal to one another, the opposite sides are also equal. Let ABC be a triangle having the angle ABC equal to the angle ACB; then will the side AB be equal to the side AC. B D A For if AB is not equal to AC, one of them must be greater than the other. Let AB be the greater, and from it cut off DB equal to AC the less, and join CD. Then, because in the triangles DBC, ACB, DB is equal to AC, and BC is common to both triangles, also, by supposition, the angle DBC is equal to the angle ACB; therefore, the triangle DBC is equal to the triangle ACB (Prop. VI.), the less to the greater, which is absurd. Hence AB is not unequal to AC, that is, it is equal to it. Therefore, if two angles, &c. Cor. Hence, every equiangular triangle is also equilateral. PROPOSITION XII. THEOREM. The greater side of every triangle is opposite to the greater angle; and, conversely, the greater angle is opposite to the greater side. Let ABC be a triangle, having the angle ACB A greater than the angle ABC; then will the side AB be greater than the side AC. Draw the straight line CD, making the angle BCD equal to B; then, in the triangle CDB, the side CD must be equal to DB (Prop. XI.). Add AD to each, then will the sum of AD and DC D C B be equal to the sum of AD and DB. But AC is less than the sum of AD and DC (Prop. VIII.); it is, therefore, less than AB. Conversely, if the side AB is greater than the side AC, then will the angle ACB be greater than the angle ABC. For if ACB is not greater than ABC, it must be either equal to it, or less. It is not equal, because then the side AB would be equal to the side AC (Prop. XI.), which is contrary to the supposition. Neither is it less, because then the side AB would be less than the side AC, according to the former part of this proposition; hence ACB must be greater than ABC. Therefore, the greater side, &c. X PROPOSITION XIII. THEOREM. If two triangles have two sides of the one equal to two sides of the other, each to each, but the included angles unequal, the base of that which has the greater angle,will be greater than the base of the other. Let ABC, DEF be two triangles, having two sides of the one equal to two sides of the other, viz.: AB equal to DE, and AC to DF, but the angle BAC greater than the angle EDF; then will the base BC be greater than the B base EF. D AA CE F G Of the two sides DE, DF, let DE be the side which is not greater than the other; and at the point D, in the straight line DE, make the angle EDG equal to BAC; make DG equal to AC or DF, and join EG, GF. Because, in the triangles ABC, DEG, AB is equal to DE, and AC to DG; also, the angle BAC is equal to the angle EDG; therefore, the base BC is equal to the base EG (Prop. VI.). Also, because DG is equal to DF, the angle DFG is equal to the angle DGF (Prop. X.). But the angle DGF is greater than the angle EGF; therefore the angle DFG is greater than EGF; and much more is the angle EFG greater than the angle EGF. Now, in the triangle EFG, because the angle EFG is greater than EGF, and because the greater side is opposite the greater angle (Prop. XII.), the side EG is greater than the side EF. But EG has been proved equal to BC; and hence BC is greater than EF. Therefore, f two triangles, &c. PROPOSITION XIV. THEOREM (Converse of Prop XIII.). If two triangles have two sides of the one equal to two sides of the other, each to each, but the bases unequal, the angle contained by the sides of that which has the greater base, will be greater than the angle contained by the sides of the other. Let ABC, DEF be two triangles having two sides of the one equal to Ꭺ two sides of the other, viz.: AB equal to DE, and AC to DF, but the base BC greater than the base EF; then will the angle BAC be greater than the angle EDF. B D E C F For if it is not greater, it must be either equal to it, or less. But the angle BAC is not equal to the angle EDF, because then the base BC would be equal to the base EF (Prop. VI.), which is contrary to the supposition. Neither is it less, because then the base BC would be less than the base EF (Prop. XIII.), which is also contrary to the supposition; therefore, the angle BAC is not less than the angle EDF, and it has been proved that it is not equal to it; hence the angle BAC must be greater than the angle EDF. Therefore, if two triangles, &c. PROPOSITION XV. THEOREM. If two triangles have the three sides of the one equal to the three sides of the other, each to each, the three angles will also · be equal, each to each, and the triangles themselves will be equal. Let ABC, DEF be two triangles having the three sides of the one equal to the three sides of the other, viz. AB equal to DE, BC to EF, and AC to DF; then will : the three angles also be equal, B viz. the angle A to the angle D, : CE D F the angle B to the angle E, and the angle C to the angle F. For if the angle A is not equal to the angle D, it must be either greater or less. It is not greater, because then the base BC would be greater than the base EF (Prop. XIII ) which is contrary to the hypothesis; neithe: is it less, be Cause then the base BC would be less than the base E (Prop. XIII.), which is also contrary to the hypothesis. Therefore, the angle A must be equal to the angle D. In the same manner, it may be proved that the angle B is equal to the angle E, and the angle C to the angle F; hence the two triangles are equal. Therefore, if two triangles, &c. Scholium. In equal triangles, the equal angles are oppo site to the equal sides; thus, the equal angles A and D are opposite to the equal sides BC, EF. PROPOSITION XVI. THEOREM. From a point without a straight line, only one perpendicular can be drawn to that line. Let A be the given point, and DE the given straight line; from the point A only one perpendicular can be drawn to DE. C. B E For, if possible, let there be drawn two perpendiculars AB, AC. Produce the line AB to F, making BF equal to AB, and join CF. Then, in the triangles ABC, FBC, because AB is equal to BF, BC is common to both triangles, and the angle ABC is equal to the angle FBC. being both right angles (Prop. II., Cor. 1); therefore, two sides and the included angle of one triangle, are equal to two sides and the included angle of the other triangle; hence the angle ACB is equal to the angle FCB (Prop. VI.). But, since the angle ACB is, by supposition, a right angle, FCB must also be a right angle; and the two adjacent angles BCA, BCF, being together equal to two right angles, the two straight lines AC, CF must form one and the same straight line (Prop. III.); that is, between the two points A and F, two straight lines, ABF, ACF, may be drawn, which is impossible (Axiom 11); hence AB and AC can not both be perpendicular to DE. Therefore, from a point, &c. Cor. From the same point, C, in the D. E line AB, more than one perpendicular to this line can not be drawn. For, if possi ble, let CD and CE be two perpendicu lars; then, because CD is perpendicular the angle ECA is also a right angle. Hence, the angle ACD is equal to the angle ACE (Prop. I.), the less to the greater,. |