3. In AB, the longest side of the triangle ABC, AD is cut off equal to AC; shew that BDC is an obtuse angle. 4. From any point outside a straight line two equal straight lines can be drawn to the line, but only two. 5. The difference of any two sides of a triangle is greater than the third side. Any two angles of a triangle are together less than two right angles. Let ABC be a triangle; then shall any two of its angles together be less than two right angles. Because ACD is the exterior angle of the triangle ABC, therefore the angle ACD is greater than the angle ABC; (prop. 16) to each of these unequals add the angle ACB; then the angles ACD, ACB are greater than the angles ABC, ACB; but the angles ACD, ACB are together equal to two right angles, (prop. 13) therefore the angles ABC, ACB are together less than two right angles. Similarly it may be shewn that the angles BAC, ACB are together less than two right angles, and also the angles BAC, ABC. Q.F.D. Exercises. 1. Shew with the above figure that the angles BAC, ACB are together less than two right angles. 2. Shew by producing AB that the angles ABC, BAC are together less than two right angles. 3. Prove the proposition without producing any of the sides, by joining A with any point in BC. 4. In an isosceles triangle shew that each of the angles at the base is less than a right angle. Do we know anything as to the magnitude of the third angle? PROPOSITION XVIII. THEOREM. If one side of a triangle is greater than another, the angle opposite the greater side is greater than the angle opposite the less. Let ABC be a triangle in which AC is greater than AB; From AC the greater cut off AD equal to AB the less. (prop. 3) Demonstration. Because BCD is a triangle of which one side CD is produced to A, therefore the exterior angle ADB is greater than the interior opposite angle ACB; (prop. 16) AD is equal to AB, but since in the triangle ADB, therefore the angle ADB is equal to the angle ABD; (prop. 5) therefore also the angle ABD is greater than the angle ACB; much more then is the angle ABC greater than the angle ACB. Q.E.D. Exercises. 1. Shew that whatever be the length of AC, the angle ABD is always greater than DBC. 2. If BC is greater than CA, shew that the angle BAC is greater than the angle ABC. 3. In a right-angled triangle ABC it is given that AB is the longest side; determine which is the right angle. 4. AD is the longest side of a quadrilateral ABCD, and BC the shortest, shew that the angle ABC is greater than the angle ADC, and the angle BCD greater than the angle BAD. D PROPOSITION XIX. THEOREM. If one angle of a triangle be greater than another, the side opposite the greater angle is greater than the side opposite the less. Let ABC be a triangle in which the angle ABC is greater than the angle ACB; then shall the side AC be greater than the side AB. A B Demonstration. If AC be not greater that AB, it must be either equal to it or less than it. then the angle ABC would be equal to the angle ACB; (prop. 5) but it is not equal to it; (hyp.) and therefore AC is not equal to AB. Again, if AC were less than AB, then the angle ABC would be less than the angle ACB; (prop. 18) but it is not less than it; (hyp.) and therefore AC is not less than AB. Hence AC is neither equal to nor less than AB; that is, AC is greater than AB. Exercises. Q.E.D. 1. In a right-angled triangle, the side opposite the right angle is the longest side. 2. Given a straight line and a point outside it; shew that of all straight lines drawn from the point to the straight line the perpendicular is the least, and that one nearer to the perpendicular is less than one more remote. 3. If ABC be an isosceles triangle, and BC be produced to D, then AD is greater than AC. 4. In AC a side of a triangle ABC, a point E is found such that BE is equal to BC; shew that AB is greater than BE. 5. If one angle of a triangle be bisected by a straight line which cuts the opposite side, each of the segments into which it divides it is less than the side of the triangle adjacent to that segment. 6. ABCD is a quadrilateral such that the diagonal AC is equal to the side AD: shew that BD is greater than BC. PROPOSITION XX. THEOREM. Any two sides of a triangle are together greater than the third side. Let ABC be a triangle ; then shall any two of its sides be greater than the third side. D B Construction. Produce BA to D, making AD equal to AC: join CD. Demonstration. In the triangle ACD, the side AC is equal to the side AD, therefore the angle ACD is equal to the angle ADC; (prop. 5) but the angle BCD is greater than the angle ACD, therefore the angle BCD is also greater than the angle ADC (i.e. BDC); therefore the side BD is greater than the side BC; (prop. 19) but BD is equal to BA, AC (because AD is equal to AC), therefore BA, AC are greater than BC. Similarly it may be shewn that AB, BC are greater than AC, and AC, CB greater than AB. Exercises. Q.E.D. 1. Construct the figure so as to prove (1) that AB, BC are greater than AC, and (2) that AC, CB are greater than AB. 2. Prove the proposition without producing one of the sides by bisecting one of the angles. 3. The four sides of any quadrilateral are together greater than twice either of the diagonals. 4. If any point be taken within a triangle, the sum of its distances from the angular points is greater than half the sum of the sides of the triangle. 5. Any three sides of a quadrilateral are together greater than the fourth side. 6. The sum of the sides of any quadrilateral is greater than the sum of the diagonals. 7. Any two sides of a triangle are together greater than twice the line drawn from the middle point of the base to the opposite angle. D If from the ends of a side of a triangle two straight lines be drawn to a point within the triangle, these shall be less than the other two sides of the triangle, but shall contain a greater angle. Let ABC be a triangle, and from B and C let straight lines BD, CD be drawn to D a point within the triangle; then shall BD, CD be less than BA, CA, but they shall contain an angle BDC greater than the angle BAC. In the triangle ABE, the two sides BA, AE are together greater than the third side BE; (prop. 20) add to each of these unequals EC; then BA, AC are together greater than BE, EC. Again, in the triangle CDE, the two sides DE, EC are together greater than the third side DC; (prop. 20) add to each of these unequals BD; then BE, EC are together greater than BD, DC. But it has been shewn that BA, AC are together greater than BE, EC; much more then are BA, AC together greater than BD, DC. Next, because the side ED of the triangle CED is produced to B, therefore the exterior angle BDC is greater than the interior opposite angle DEC; (prop. 16) similarly, because the side AE of the triangle BAE is produced to C; therefore the exterior angle BEC is greater than the interior opposite angle BAC : much more then is the angle BDC greater than the angle BAC. Q. E.D. |