Proposition III. From the greater of two given straight lines a part may be cut off equal to the less. Postulate 1. A straight line may be drawn from any one point to any other point. Proposition IV. If two triangles have two sides of the one equal to two sides of the other, each to each, and have also the angles contained by those sides equal to one another the two triangles shall be equal. Axiom 9. The whole is greater than its part. PROPOSITION VII. THEOREM. On the same base, and on the same side of it, there cannot be two triangles having their sides which are terminated at one extremity of the base equal to one another, and likewise those which are terminated at the other extremity equal to one another. In the two triangles ABC and ABD; GIVEN AC equal to AD, and BC equal to BD; IT IS REQUIRED TO PROVE that the triangles cannot be on the same base AB and on the same side of it, without coinciding. Suppose that the triangles can be thus situated; First, with the vertex of each triangle without the other triangle, Because AC is equal to AD; therefore the angle ACD is equal to the angle ADC. Because the angle ACD is greater than the angle BCD, Because the angle BDC is greater than the angle ADC, and the angle ADC is greater than the angle BCD; therefore the angle BDC is greater than the angle BCD. Because BC is equal to BD; Post. 1. Нур. I. 5. Axiom 9. Proved. t. Axiom 9. therefore the angle BDC is equal to the angle BCD. Because, if the vertex of each triangle is without the other triangle, then the angle BDC is both equal to and greater than the angle BCD, and this is impossible; V. Нур. I. 5. Proved. therefore the vertex of each triangle cannot be without the other triangle. n. G. Postulate 1. A straight line may be drawn from any one point to any other point. Proposition V. The angles at the base of an isosceles triangle are equal to one another. Axiom 9. The whole is greater than its part. Produce AC and AD to E and F. Because AC is equal to AD; therefore the angle ECD is equal to the angle FDC. Because the angle ECD is greater than the angle BCD, Because the angle BDC is greater than the angle FDC, and the angle FDC is greater than the angle BCD; therefore the angle BDC is greater than the angle BCD. Because BD is equal to BC; therefore the angle BDC is equal to the angle BCD. Because if D be within the triangle ACB, Post. 1. Нур. I. 5. Axiom 9. t. Axiom 9. then the angle BDC is both equal to and greater than the angle BCD, and this is impossible; therefore D cannot be within the triangle ACB. The third and only other case; V. Нур. I. 5. Proved. in which the vertex of one triangle is on a side of the other, needs no demonstration. n. G. Therefore, on the same base &c. Q. E. D. Postulate 1. A straight line may be drawn from any point to any other point. Postulate 2. A terminated straight line may be produced to any length in a straight line. Proposition V. The angles at the other side of the base of an isosceles triangle are equal to one another. Axiom 9. The whole is greater than its part. |