SECTION III. TRIANGLES. Def. 22. A triangle is the figure contained by three straight lines, which are called its sides. The sum of the sides is sometimes called the perimeter. Def. 23. sides are equal. Def. 24. A triangle is called isosceles when two of its A triangle is called equilateral when all its sides are equal, and equiangular when all its angles are equal. Def. 25. It is said to be right-angled or obtuse-angled when one of the angles is a right angle or an obtuse angle, and acute-angled when all its angles are acute angles. Def. 26. A triangle is sometimes regarded as standing on one of its sides, which is then called its base; and the angle opposite that side is called the vertex. When two of the sides of a triangle have been mentioned, the remaining side is often spoken of as the base. Def. 27. The term hypothenuse is used to describe the side of a right-angled triangle which is opposite to the right angle. Def. 28. The area of a figure is the space enclosed by the sides of the figure. Def. 29. Triangles are said to be equal in all respects when they have the three sides and the three angles of the one equal to the three sides and the three angles respectively of the other, and the area of one equal to the area of the other. Ax. 10. A triangle may be conceived as taken up and placed in any other position, the magnitude of its parts remaining unaltered. THEOREM 7. If one side of a triangle is produced, the exterior angle will be equal to the two interior and opposite angles, and the three interior angles of any triangle will be together equal to two right angles. B C Let the side BC of the triangle ABC be produced to D: then shall the angle ACD = the sum of the angles ABC, CAB. And the three angles ABC, BCA, CAB shall be together equal to two right angles. Proof. For if through C a line CH were drawn parallel to BA, the angle HCD= the corresponding angle ABC (Th. 4), and the angle ACH the alternate angle BAC (Th. 4); .. the whole angle ACD = the two angles ABC+BAC. And if ACB be added to these, the two angles ACD+ACB= the three angles ABC+ ВСА + САВ. But ACD+ACB= two right angles (Th. 1); therefore ABC+BCA + CAB = two right angles. Remark. This latter result is obviously a particular case of the preceding Theorem. COR. I. It follows that no triangle can have more than one right angle or obtuse angle. COR. 2. In a right-angled or obtuse-angled triangle the right or obtuse angle is the greatest angle. COR. 3. In any right-angled triangle the two acute angles together make up one right angle. COR. 4. An exterior angle of a triangle is greater than either of the interior and opposite angles. The four next Theorems are properties of a single triangle. THEOREM 8. An isosceles triangle will have the angles at its base equal. Let ABC be an isosceles triangle, having AB = AC. B A Then shall the angle B = the angle C. Proof. For conceive the triangle ABC taken up and put down in its former position, with the arms of the angle A transposed. (Ax. 10.) That is, let A be put down where it was before, and AC where AB was; then AB would be where AC was, since the angle at A is unchanged, and B would fall where C was, and C where B was, because AB=AC; and BC would fall as it was before, but in a reversed position; that is, the angle B would fall on the angle C, and therefore the angle B = the angle C. COR. I. The angles on the other side of the base, made by producing the equal sides, will be equal. For they are respectively supplementary to the angles at the base (Th. 1). COR. 2. An equilateral triangle will be equiangular. THEOREM 9. Oppositely. A triangle which is not isosceles will have the angles at its base unequal, the greater angle being opposite to the greater side. is greater than AC. Let ABC be a triangle in which AB Then will the angle ACB be greater than the angle ABC. Proof. Conceive the triangle BAC taken up and put down with the arms of and the angle at A transposed: that is, let AC fall as AC", AB as AB, and therefore BC as B'C'; and let BC, B'C' intersect in O. Then the angle ACB is the same as the angle AC'B; but AC'B' is greater than ABC, because it is the exterior angle of the triangle OC'B (Th. 7, Cor. 4); and therefore ACB is greater than ABC. THEOREM IO. Conversely. A triangle which has the angles at its base equal will be isosceles. Let the triangle ABC have the angles B and C equal. Then shall AB= AC. B Proof. For AB is not unequal to AC, for then the angle C would be unequal to the angle B (Th. 9), which it is not and therefore AB = AC. COR. An equiangular triangle is equilateral. THEOREM II. Conversely to the opposite. A triangle which has two of its angles unequal shall have the sides opposite them unequal, the greater side being opposite to the greater angle. Let the angle ACB be greater than the angle ABC, then shall AB be greater than AC. Proof. For AB is not equal to AC, for then (by Th. 8) the angle ABC would be equal to the angle ACB. B Nor is AB less than AC, for then the angle ACB would be less than ABC (by Th. 9); Therefore AB is greater than AC. A C THEOREM 12. Of all the straight lines that can be drawn from a given point to meet a given straight line, the shortest will be the perpendicular; and of the others, that which is further from the perpendicular will be greater than that which is nearer; and one, and only one, oblique can be drawn equal to any given oblique, and it will be on the other side of the perpendicular and equally inclined to it. Let OQ be any other line from O meeting AB in Q; OP shall be less than OQ. Proof. For since OPQ is a right angle, OQP is an acute angle (Th. 7, Cor. 2); that is, OPQ> OQP, and therefore OQ> OP by Theorem 11. Hence OP is less than any other line OQ, and therefore is the least of all the lines drawn from 0 to AB. Next let OQ, OR be two obliques of which OR is the further from the perpendicular, that is, makes the greater angle with it; then OR shall be greater than OQ. Proof. For since OPQ is a right angle, OQR the exterior angle is an obtuse angle (Th. 2, Cor. 4); and ORQ is an acute angle; therefore OQR is > ORQ, and therefore OR is > OQ (Theorem 11). Lastly, there can be drawn one, and only one, oblique equal to a given oblique, which will be equally inclined to the perpendicular and on the opposite side of it. Proof. For if PS=PR, and the figure be conceived as folded on OP, it is clear that since the angles at P are right. |