THEOREM II. In any triangle the square on the side opposite an acute angle is less than the squares on the other two sides by twice the rectangle contained by either side and the projection on it of the other side *. Part. En. Let ABC be a triangle, B an acute angle, BD the projection of AB on BC, then will AC = AB® + BC? – 2CB x BD. • Cor. Conversely, the angle opposite a side of a triangle is an acute angle, a right angle, or an obtuse angle, according as the square on that side is less than, equal to, or greater than the sum of the squares on the other two sides. THEOREM 12. The sum of the squares on two sides of a triangle is double the sum of the squares on half the base and on the line joining the vertex to the middle point of the base. * Euclid, 11. 13. Part. En. Let AC, a side of the triangle ABC, be bisected in D; then will AB’ + BC= 2 ADP + 2BD'. Proof. For let DE be the projection L U O of BD on AC. Then ABP = ADP + DBR + 2AD. DE (by 11. 10), and BC = CD + DB® – 2CD. ED (by 11. 11), therefore remembering that AD=DC, we obtain by addition that AB® + BCP= 2 ADP +2DBR. This theorem in a more general form is known as Apollonius's Theorem. THEOREM 13. If a straight line is divided internally or externally at any point, the sum of the squares on the segments is double the sum of the squares on half the line and on the line between the point of division and the middle point of the line*. Let AB be bisected in C, and divided internally or externally in D. A D B La Then the squares on AD, DB will be double of the squares on AC, CD. Proof. For ADP = AC? + CD +2 ACx CD by 11. 6; and DBP = CBR + CD – 2BCR CD by 11. 7; therefore, adding, and remembering that AC=BC, and that therefore the rectangle AC CD)= the rectangle BC CD, we get that AD + DBP = 2 AC' + 2 CD', * Eucl. 11. 9, 10. EXERCISES. 1. Bisect a triangle by a line passing through one of its angular points. 2. Any line drawn through the intersection of the diagonals of a parallelogram to meet the sides bisects the figure. . . . . . . . . 3. Find the locus of the vertices of triangles of equal area upon the same base. 4. If the sides of a triangle are 3, 4, 5 inches respectively, the triangle is right-angled. 5. Of all triangles having the same vertical angle, and whose bases pass through a given point, the least is that whose base is bisected in that point. 6. The diagonals of a parallelogram divide it into four equivalent triangles. 7. If from any point in the diagonal of a parallelogram: straight lines be drawn to the angles, then the parallelogram will be divided into two pairs of equivalent triangles. 8. ABCD is a parallelogram, and E any point in the diagonal AC produced. Shew that the triangles EBC, EDC will be equivalent. . 9. ABCD is a parallelogram, and 0 any point within it, shew that the triangles OAB, OCD are together equivalent to half the parallelogram... 10. On the same supposition if lines are drawn through O parallel to the sides of the parallelogram, then, the difference of the parallelograms DO, BO is double of the triangle OAC... w. will be unes be drawn to in the diagonal of II. The diagonals of a parallelogram ABCD intersect in O, and P is a point within the triangle OAB. Prove that the difference of the triangles APB, CPD, is equivalent to the sum of the triangles APC, BPD. 12. If the points of bisection of the sides of a triangle be joined, the triangle so formed shall be one-fourth of the given triangle. 13. Shew that the sum of the squares on the lines joining the angular points of a square to any point within it is double of the sum of the squares on the perpendiculars from that point on the sides. 14. If the sides of a quadrilateral figure be bisected, and the points of bisection joined, prove that the figure so formed will be a parallelogram equal in area to half the given quadrilateral. 15. Bisect a parallelogram by a line passing through any given point. SECTION II. PROBLEMS. On the Quadrature of a Rectilineal Area. There is one problem which from its historical interest, and from the valuable illustrations it affords of the methods and limitations of Geometry, should find a place there. This problem is called the quadrature of a retilineal area, which means the finding a square whose area is equivalent to that of any given figure which is bounded by straight lines. It gave a means of comparing any two dissimilarly shaped rectilineal figures, such as irregularly shaped fields whose boundaries were straight. In the present condition of mathematics it is not necessary, as the student will hereafter learn, but it will always be instructive. The problem is approached by the following stages: (1) To construct a parallelogram, with sides inclined at a given angle, equal to a given triangle. (2) To construct on a given straight line a parallelogram, with sides inclined at a given angle, equal to a given triangle. (3) To construct a parallelogram, with sides inclined at a given angle, equal to a given rectilineal figure. (4) To construct a square equal to a given rectilineal figure. A PROBLEM I. To construct a parallelogram equal to a given triangle and having one of its angles equal to a given angle. Let ABC be the given tri- B ---angle, E the given angle. Construction. Bisect AC in D, make the angle CDF=E, and through B draw BFH parallel to AC, and draw CH parallel to DF. FDCH will be the parallelogram required. Proof. If BD be joined, it will be clear that the triangle BAC and the parallelogram FHCD are each of them double of the triangle BDC (11. 2, Cor. 1), and therefore the parallelogram FHCD= the triangle BAC, and it has an angle = E, which was required* * Euclid, 1. 42. |