angle A to the angle D, and the area ABC to the area DEF. Q. E. D. THEOREM 8. If the angles at the base of a triangle are equal to one another, the triangle is isosceles*. Part. En. Let the two angles B and C of the triangle ABC be equal; it is required to prove that AB= AC. Proof. If the triangle were taken up and reversed and replaced, so that the point C fell where B A B was, and the line CB along the line BC, then B would fall where C was. And because the angle C the angle B, = (Hyp.) the line CA would lie along BA, and BA along CA; therefore the point A would coincide with its former position, and the lines AC, AB would coincide with the lines AB, AC. Therefore AB = AC Q. E. D. COR. If a triangle is equiangular, it is also equilateral. THEOREM 9. If any side of a triangle be produced, the exterior angle will be greater than either of the interior and opposite angles. Part. En. Let ABC be a triangle, and let one of its sides BC be produced to D; it is required to prove that the exterior angle ACD is greater than either of the interior and opposite angles CAB or ABC. * Euclid, 1. 6. Proof. Firstly to prove that ACD is greater than BAC. Let AC be bisected in E. (Ax. 3.) E C Join BE, and produce it to F, making EF=EB. And join FC. Then in the triangles AEB, CEF we have AE= EC, (Constr.) (Constr.) and the contained angles AEB, CEF are equal; (Th. 4.), therefore the triangles are equal in all respects; and therefore the angle EAB= the angle ECF. (Th. 5.) But the angle ECD is greater than ECF; therefore the angle ECD is also greater than EAB. Again, if AC is produced to G, and BC is bisected, it may be similarly shewn that BCG is greater than ABC; but BCG is equal to ACD, therefore ACD is also greater than ABC; that is, the exterior angle ACD is greater than either CAB or ABC*. Q. E. D. THEOREM IO. The greater side of every triangle has the greater angle opposite to it. * Euclid, 1. 16. Part. En. Let ABC be a triangle having AC greater than AB; B с it is required to prove that the angle ABC is greater than the angle ACB. Proof. From AC cut off AD = AB; and join DB. AD = AB; therefore the angle ABD = the angle ADB. (Constr.) (Th. 6.) But because ADB is the exterior angle of the triangle BDC, therefore the angle ADB is greater than the angle ACB; (Th. 9.) 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. THEOREM II. The greater angle of every triangle has the greater side opposite to it. Part. En. Let ABC be a triangle in which the angle B is greater than the angle C; it is required to prove that the side AC B is greater than the side AB. C Proof. For AC must be either equal to AB, or less than AB, or greater than AB. *Euclid, I. 18. But AC is not equal to AB, for then the angle B would be equal to the angle C. Nor is AC less than AB, (Th. 6.) for then the angle B would be less than the angle C. (Th. 10.) Therefore AC is greater than AB*. Q. E. D. THEOREM 12. Any two sides of a triangle are together greater than the third side. Part. En. Let ABC be a triangle; it is required to prove that AB and BC are together greater than AC. B Proof. Produce AB to D, making BD=BC; join DC. Then because BD=BC, therefore the angle BCD = the angle BDC. (Constr.) (Th. 6.) But the angle ACD is greater than the angle BCD; therefore the angle ACD is greater than the angle ADC; and therefore AD is greater than AC. (Th. 11.) But AD is equal to AB and BC together, therefore AB and BC are together greater than AC†. Q.E.D. COR. The difference of any two sides of a triangle is less than the third side. THEOREM 13. If from the ends of the 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. Part. En. Let ACB be a triangle, and from the ends of the side AB let two straight lines AP, BP be drawn to a point P within the triangle; it is required to prove that AP and PB are less than AC and CB, but the angle APB greater than the angle ACB. Proof. Produce AP to meet BC in Q. Because any two sides of a triangle are together greater than the third side, (Th. 12.) therefore AC and CQ are greater than AQ; add to each QB; A B therefore AC and CB are greater than AQ and QB. Again, because PQ and QB are greater than PB; add to each AP; (Th. 12.) therefore AQ and QB are greater than AP and PB ; but AC and CB are greater than AQ and QB ; much more then are AC and CB greater than AP and PB. Again, because APB is the exterior angle of the triangle PQB: therefore the angle APB is greater than the angle PQB; (Th. 9.) and because PQB is the exterior angle of the triangle ACQ; therefore the angle PQB is greater than the angle ACQ; (Th. 9.) |