4. ABC is an isosceles triangle, having AB = AC. AD drawn to the base BC does not bisect A; prove that D is at unequal distances from B and C. 5. Prove the proposition with the same construction as in the text, but let ▲ DEG fall on the other side of DE. If two triangles have two sides of the one respectively equal to two sides of the other, but their bases unequal, the angle contained by the two sides of the triangle which has the greater base shall be greater than the angle contained by the two sides of the other. да CE F Let ABC, DEF be two triangles, having AB = DE, AC = DF, but base BC greater than base EF: it is required to prove ▲ A greater than ▲ D. If A be not greater than D, it must be either equal to D, or less than D. But = A is not = LD, for then base BC would be base EF, which it is not. And A is not less than D, for then base BC would be less than base EF, which it is not. .. A must be greater than ▲ D. I. 4 Hyp. I. 24 Hyp. 1. In the figure to the first deduction on I. 24, if AB is greater than BC, prove that AOB is greater than BOC. 2. ABCD is a quadrilateral, having AB = CD, but the diagonal BD greater than the diagonal AC; prove that DCB is greater than 4 ABC. 3. ABCD is a quadrilateral, having AB = CD, but ▲ BCD greater than ABC; prove that DAB is greater than ADC. 4. ABCD is a quadrilateral, having AB = CD, but 2 DAB greater than ADC; prove that BCD is greater than ▲ ABC. Dis the middle 5. ABC is a triangle, having AB less than AC. point of BC, and AD is joined; prove that 6. ABC is an isosceles triangle, having AB: = ADB is acute. AC. Dis any point such that BD is greater than DC; prove that AD does not bisect A. 7. ABC is a triangle, having AB less than AC, and AD is the median drawn from A; prove that G, any point in AD, is nearer to B than to C. If two angles and a side in one triangle be respectively equal to two angles and the corresponding side in another triangle, the two triangles shall be equal in every respect; that is, (1) The remaining sides of the one triangle shall be equal to the remaining sides of the other. (2) The third angles shall be equal. (3) The areas of the two triangles shall be equal. L it is required to prove AB = DE, AC = DF, ▲ A = LD, If AB be not = DE, one of them must be the greater. Let AB be the greater, and make BG = DE; I. 3 Const. Hyp. Нур. .. GCB = = ▲ ACB, which is impossible. Hence AB is not unequal to DE, that is, AB AB = DE Proved .. AC = DF, L A = L D, ▲ ABC CASE 2. D = да B H E = LE, L C F = F, and If BC be not = EF, one of them must be the greater. Let BC be the greater, and make BH and join AH. = EF; E I. 3 = ... LAHB L ACB, which is impossible. Hence BC is not unequal to EF, that is, BC AB == DE Now in As ABC, DEF, BC = EF LB = LE; Proved Нур. .. AC = DF, ▲ BAC = ▲ EDF, ▲ ABC = ▲ DEF. I.4 1. Prove the first case of the proposition by superposition. 2. The straight line that bisects the vertical angle of an isosceles triangle bisects the base, and is perpendicular to the base. 3. The straight line drawn from the vertical angle of an isosceles triangle perpendicular to the base, bisects the base and the vertical angle. 4. Any point in the bisector of an angle is equidistant from the arms of the angle. 5. In a given straight line, find a point such that the perpendiculars drawn from it to two other straight lines may be equal. 6. Through a given point, draw a straight line which shall be equidistant from two other given points. 7. Through a given point, draw a straight line which shall form with two given intersecting straight lines an isosceles triangle. If two sides of one triangle be respectively equal to two sides of another triangle, and if the angles opposite to one pair of equal sides be equal, the angles opposite the other pair of equal sides shall either be equal or supplementary. In As ABC, DEF let AB LE: = DE, AC = DF, ▲ B = it is required to prove either C = L F, or L C+ LF .. As ABC, DEF are equal in all respects, and LCLF. Hyp. I. 26 |