PROPOSITION XII. THEOREM. In obtuse-angled triangles, if a perpendicular be drawn from either of the acute angles to the opposite side produced, the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle, by twice the rectangle contained by the side, upon which, when produced, the perpendi cular falls, and the straight line intercepted without the triangle between the perpendicular and the obtuse angle. Let ABC be an obtuse-angled ▲, having ▲ ACB obtuse. Then must sq. on AB be greater than sum of sqq. on BC, CA by twice rect. BC, CD. For since BD is divided into two parts in C, sq. on BD=sum of sqq. on BC, CD, and twice rect. BC, CD. Add to each sq. on DA: then II. 4. sum of sqq. on BD, DA=sum of sqq. on BC, CD, DA and twice rect. BC, CD. I. 47. I. 47. Now sqq. on BD, DA=sq. on AB, and sqq. on CD, DA=sq. on CA; .. sq. on AB sum of sqq. on BC, CA and twice rect. BC, CD. .. sq. on AB is greater than sum of sqq. on BC, CA by twice rect. BC, CD. Q. E. D. Ex. The squares on the diagonals of a trapezium are together equal to the squares on its two sides, which are not parallel, and twice the rectangle contained by the sides, which are parallel. In every triangle, the square on the side subtending any of the acute angles is less than the squares on the sides containing that angle, by twice the rectangle contained by either of these sides and the straight line intercepted between the perpendicular, let fall upon it from the opposite angle, and the acute angle. FIG. 1. FIG. 2. B C Let ABC be any ▲, having the ▲ ABC acute. Then must sq. on AC be less than the sum of sqq. on AB, BC, by twice rect. BC, BD. For in Fig. 1 BC is divided into two parts in D, and in Fig. 2 BD is divided into two parts in C; .. in both cases sum of sqq. on BC, BD=sum of twice rect. BC, BD and sq. on CD. Add to each the sq. on DA, then II. 7. sum of sqq. on BC, BD, DA=sum of twice rect. BC, BD and sqq. on CD, DA ; .. sum of sqq. on BC, AB=sum of twice rect. BC, BD and sq. on AC; I. 47. .. sq. on AC is less than sum of sqq. on AB, BC by twice rect. BC, BD. The case, in which the perpendicular AD coincides with AC, needs no proof. Q. E. D. Ex. Prove that the sum of the squares on any two sides of a triangle is equal to twice the sum of the squares on half the base and on the line joining the vertical angle with the middle point of the base, PROPOSITION XIV. PROBLEM. To describe a square that shall be equal to a given rectilinear figure. Let A be the given rectil. figure. Then if BE-ED the I. 45. BCDE is a square, and what was reqd. is done. But if BE be not=ED, produce BE to F, so that EF-ED. Bisect BF in G; and with centre G and distance GB, describe the semicircle BHF. Produce DE to H and join GH. Then, BF is divided equally in G and unequally in E, .'. rect. BE, EF together with sq. on GE =sum of sqq. on EH, GE. Take from each the square on GE. Then rect. BE, EF=sq. on EH. But rect. BE, EF=BD, :: EF=ED; .. sq. on EH=BD; .. sq. on EH=rectil. figure A. II. 5. I. 47. Q. E. F. Miscellaneous Exercises on Book II. 1. In a triangle, whose vertical angle is a right angle, a straight line is drawn from the vertex perpendicular to the base; shew that the square on either of the sides adjacent to the right angle is equal to the rectangle contained by the base and the segment of it adjacent to that side. 2. The squares on the diagonals of a parallelogram are together equal to the squares on the four sides. 3. If ABCD be any rectangle, and O any point either within or without the rectangle, shew that the sum of the squares on OA, OC is equal to the sum of the squares on OB, OD. 4. If either diagonal of a parallelogram be equal to one of the sides about the opposite angle of the figure, the square on it shall be less than the square on the other diameter, by twice the square on the other side about that opposite angle. 5. Produce a given straight line AB to C, so that the rectangle, contained by the sum and difference of AB and AC, may be equal to a given square. 6. Shew that the sum of the squares on the diagonals of any quadrilateral is less than the sum of the squares on the four sides, by four times the square on the line joining the middle points of the diagonals. 7. If the square on the perpendicular from the vertex of a triangle is equal to the rectangle, contained by the segments of the base, the vertical angle is a right angle. 8. If two straight lines be given, shew how to produce one of them so that the rectangle contained by it and the produced part may be equal to the square on the other. 9. If a straight line be divided into three parts, the square on the whole line is equal to the sum of the squares on the parts together with twice the rectangle contained by each two of the parts. 10. In any quadrilateral the squares on the diagonals are together equal to twice the sum of the squares on the straight lines joining the middle points of opposite sides. 11. If straight lines be drawn from each angle of a triangle to bisect the opposite sides, four times the sum of the squares on these lines is equal to three times the sum of the squares on the sides of the triangle. 12. CD is drawn perpendicular to AB, a side of the triangle ABC, in which AC=AB. Shew that the square on CD is equal to the square on BD together with twice the rectangle AD, DB. 13. The hypotenuse AB of a right-angled triangle ABC is trisected in the points D, E; prove that if CD, CE be joined, the sum of the squares on the sides of the triangle CDE is equal to two-thirds of the square on AB. 14. The square on the hypotenuse of an isosceles right-angled triangle is equal to four times the square on the perpendicular from the right angle on the hypotenuse. 15. Divide a given straight line into two parts, so that the rectangle contained by them shall be equal to the square described upon a straight line, which is less than half the line divided. |