2. Given the sum of the three sides, and the angles at the base of a triangle, to construct it. Let AB be the sum of the three sides, and c and D the two given angles: It is required to describe a triangle, which shall have the sum of its sides equal to AB, and the angles at the base equal to c and D respectively. At the point A make (E. 1. 23.) the angle BAE equal to the angle c, and at the point в the angle ABK equal to the angle D. Bisect (E. 1.9.) the angles BAE and ABK, by the lines AF and BF, meeting in F; through F draw (E. I. 31.) FG parallel to EA, and FH parallel to KB; then GHF is the triangle required. For since FG is parallel to EA, therefore (E. 1. 29.) the angle EAF is equal to the angle GFA; but (by construction) the angle EAF is equal to the angle GAF; therefore the angle GAF is equal to the angle GFA; therefore (E. 1. 6.) GF is equal to GA. In the same manner it may be shown that HF is equal to HB; hence the sum of the three sides of the triangle GHF is equal to the given straight line AB. Again, since FG is parallel to EA, and FH to KB, therefore (E. I. 29.) the angle FGH is equal to the angle EAB, and the angle FHG to the angle KBA; but (by construction) the angle EA B is equal to the angle C, and the angle KBA is equal to the angle D; therefore also the angle FGH is equal to the angle C, and the angle FHG to the angle D. Wherefore, &c. 3. To divide a given straight line into any given number of equal parts. Let CB be a given straight line it is required to divide it into any given number of equal parts. A m n p B From c draw an indefinite line CA making any convenient angle with CB; take any point e in CA, and make er, rs, sa, &c., each equal to ce, so that the number of these equal parts on CA may be equal to the given number of parts into which CB is to be divided. Join the extremity A of the last part with the extremity в of the given line; and through the points e, r, s, &c. draw (E. 1. 31.) eh, rm, sp, &c., parallel to A B, cutting CB in the points h, m, p, &c.; then these points will divide CB into the required number of equal parts. Draw hn, mo, parallel to CA. Because rm is parallel to eh (E. 1. 30.) and hn to er, therefore ehnr is a parallelogram, and (E. 1. 34.) hn is equal to er or ce; the angle nhm is equal to the ech, and the angle hmn to the angle che (E. 1. 29.); therefore (E. I. 26.) the triangle nhm is equal to the triangle ech, and hm is equal to ch. In like manner it may be proved that mp is equal hm, and so on. 4. To bisect a triangle by a line drawn from a given point in one of the sides. Let ABC be the triangle, and P the given point. Bisect BC in E; join AE, PE; from A draw AF parallel to PE; and join PF; then PF will bisect the triangle ABC. Since AF is parallel to B A .E F ୯ PE, therefore (E. 1. 37.) the triangle PEA is equal to the triangle PEF. To these equals add the triangle BEP, therefore the triangle ABE is equal to the triangle BPF; but since BE is equal to EC, therefore (E. 1. 38.) the triangle ABE is equal to half the triangle ABC; therefore the triangle BPF is also equal to half the triangle ABC. 5. Given the base, the sum of the remaining sides, and one of the angles at the base of a triangle, to construct it. 6. Given the base and the vertical angle of an isosceles triangle, to construct it. 7. To construct a triangle from three parts given, of which one, at least, is a side. 8. Upon a given base to describe an isosceles triangle equal to a given rectangle. 9. To divide a triangle into any given number of equal parts by lines drawn from the vertex to the base. 10. Describe a square which shall be equal to the difference between two given squares. 11. Describe a square which shall be equal to the sum of three given squares. 12. Construct a triangle which shall have the magnitude of its angles as the numbers 1, 2, and 3. 13. To find a point within a triangle, so that lines drawn to the angles shall divide the triangle into three equal parts. 14. Draw a line EF parallel to the base BC of a triangle ABC, so that EF shall be equal to BE. 15. Construct a parallelogram, having given the diagonals and the angle they make with each other. 16. Trisect a given angle, which is the half, or the quarter, or the eighth part, and so on, of a right angle. 17. In a given square to inscribe an equilateral triangle, having one of its angular points upon one of the angular points of the square, and its two remaining angular points one in each of two adjacent sides of the square. 18. To bisect a parallelogram by a line drawn through a given point in one of its sides. 19. To bisect a trapezium (1st) by a line drawn from one of its angular points, (2d) by a line drawn from a given point in one of its sides. 20. A plane rectilineal figure of any number of sides being given, to find an equal rectilineal figure which shall have the number of its sides less, by one, than that of the given figure. 21. To trisect a parallelogram by lines drawn from a given point in one of its sides. 22. To describe a parallelogram, the surface and perimeter of which shall be respectively equal to the surface and perimeter of a given triangle. 23. To describe a parallelogram which shall be of a given altitude, and equiangular with a given parallelogram, and also equal to it. 24. From the circumference of a given circle to draw to a straight line given in position, a line which shall be equal and parallel to a given straight line. 25. To describe a triangle which shall be equal to a given equilateral pentagon, and of the same altitude. 26. To inscribe a square in a given right-angled isosceles triangle. 27. To inscribe a square in a given equilateral four-sided figure. 28. To inscribe a square in a given quadrant of a circle. 29. Given one angle, a side opposite to it, and the difference of the other two sides, to construct the triangle. 30. Given the perimeter and the vertical angle of an isosceles triangle, to construct it. 31. From one of the angles of a parallelogram to draw a line to the opposite side, which shall be equal to that side together with the segment of it which is intercepted between the line and the opposite angle. 32. Two straight lines are given in position, without being produced to meet; it is required to draw a line which would (if produced) bisect the angle between the two straight lines. 33. Given the vertical angle, the difference of the angles at the base, and the perpendicular height of a triangle, to construct it. 34. In a given triangle to construct an equilateral parallelogram, which shall have one of its angles coinciding with one of the angles of the triangle, and the opposite angular point situated in one of the sides of the triangle. THEOREMS. No. II. 1. The line joining the vertex and the middle of the base of a triangle, bisects every line that is drawn parallel to the base, and is terminated by the two remaining sides of the triangle. A Let PG be any straight line, drawn parallel to the base BC of the triangle ABC; and let AD, joining the vertex and the middle D of BC, cut PQ in G: then PQ is bisected in the point G. If PG be not equal to GQ, one of them is the greater: let PG be the greater; and join DQ, and DP. Since (hyp.) BD is equal to DC, and PQ is parallel to B P F E D BC, therefore (E. 1. 38.) the triangle BDP is equal to M. |