Then the square of AB is equal to the squares of BD, DA together with twice the rectangle BD, DE, (II. 12.) and the square of AC is equal to the squares of CD, DA diminished of square BD, and twice the square of AD; for DC is equal to BD: and twice the squares of AB, AC are equal to the square of BC, and four times the square of AD: for BC is twice BD. Similarly, twice the squares of AB, BC are equal to the square of AC, and four times the square of BE: and twice the squares of BC, CA are equal to the square of AB, and four times the square of FC: hence, by adding these equals, four times the squares of AB, AC, BC are equal to four times the squares of AD, BE, CF together with the squares of AB, AC, BC: and taking the squares of AB, AC, BC from these equals, therefore three times the squares of AB, AC, BC are equal to four times the squares of AD, BE, CF. PROBLEM I. Divide a given straight line into two such parts, that the rectangle contained by them may be three-fourths of the greatest which the case will admit. Analysis. Let AB be the given line, and let AB be bisected in D: then the rectangle AD, DB, or the square of DB is the greatest possible rectangle. Let C be the point required, such that the rectangle AC, CB is equal to three-fourths of the square of DB. On AB describe a semicircle, and draw CE perpendicular to AB, then the square of CE is equal to the rectangle AC, CB. Again, bisect DB in F; on FB describe the right-angled triangle FBG, having the hypothenuse FG equal to DB. Then BG is the line, the square of which is three-fourths of the square of DB. Hence GB is equal to EC: join GE. Therefore the point E, and also the point C is found. Synthesis. Bisect AB in D and DB in F: on FB make the rightangled triangle FBG having the hypothenuse FG equal to DB. On AB describe a semicircle, and through G draw GE parallel to AB, and draw EC parallel to GB. Then C is the point required, such that the rectangle contained by AC, CB is equal to three-fourths of the square of half the line AB. PROBLEMS. 2. Divide a straight line into two parts, such, that their rectangle may be equal to a given square; and determine the greatest square that the rectangle can equal. 3. Divide a straight line into two such parts that the difference of the squares of the two parts shall be equal to twice the rectangle contained by them. 4. Divide a straight line into two parts such, that the rectangle contained by them may be equal to the square of their difference. 5. To divide a straight line so that the rectangle under its segments may equal a given rectangle. 6. Divide a given straight line so that the rectangle under the parts may be equal to a given square, and point out the limit which the side of the given square must not exceed so that the problem may be possible. 7. Divide a given straight line into two parts, such that the squares of the whole line and one of the parts shall be equal to twice the square of the other part. 8. Divide a given line, so that the square of the greater part may equal twice the rectangle of the whole and the less part. 9. Divide algebraically a given line (a) into two parts, such that the rectangle contained by the whole and one part may be equal to the square of the other. Deduce Euclid's construction from one solution, and explain the other. 10. Divide a given straight line into three parts, such that the square of the whole line may be equal to the squares of the extreme parts together with twice the rectangle contained by the whole and the middle part. 11. To produce a 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. 12. If a straight line be divided into any two parts, produce it so that the rectangle contained by the whole line produced and the part produced, may be equal to the rectangle contained by the given line and one segment. 13. Construct a rectangle that shall be equal to a given square and the difference of whose adjacent sides shall be equal to a given line. 14. Construct a rectangle equal to a given square, and having the sum of its sides equal to a given line. 15. Find a square which shall be equal to the sum of two given rectilineal figures. 16. Shew how to divide a given rectangle into parts which together will form a rectangle of any proposed length. 17. A given line AB is divided into two parts in the point C. Find the position of any point D above the line, so that the sum of the squares on the lines DA, DC, DB may be equal to the square on AB, diminished by the rectangle AC, AB. 18. Construct a triangle with three sides a, b and c, such that ab=c2, and a+b= 4c. THEOREMS. 3. The area of a rhombus is equal to half the rectangle contained by the diagonals. 4. Prove that the area of a trapezium whose bases are parallel is half of the rectangles contained by each of the bases, and the perpendicular distance. 5. The area of any right-angled triangle is equal to the rectangle of the semiperimeter and excess of the semiperimeter above the hypothenuse. Required proof. 6. Any rectangle is the half of the rectangle contained by the diameters of the squares on its two sides. 7. The sum of the squares of two lines is never less than twice their rectangle. 8. If a straight line be divided into two equal and into two unequal parts, the squares of the two unequal parts are equal to twice the rectangle contained by the two unequal parts, together with four times the square of the line between the points of section. 9. If the points C, D be equidistant from the extremities of the straight line AB, shew that the squares constructed on AD and AC exceed twice the rectangle AC, AD by the square constructed on CD. 10. In a right-angled triangle, the square on that side which is the greater of the two containing the right angle is equal to the rectangle by the sum and difference of the other sides. 11. Shew that the first of the algebraical propositions, (a + x) (a-x) + x2 = a2, (a + x)+(a-x)2 = 2a2+2x2, is equivalent to the two Propositions v and vi, and the second of them to the two Propositions IX and x of the second book of Euclid. 12. Shew how in all the possible cases, a straight line may be geometrically divided into two such parts, that the sum of their squares shall be equal to a given square. 13. ABCD is a rectangular parallelogram, of which_A, C are opposite angles, E any point in BC, F any point in CD. Prove that the area of the triangle AEF together with the rectangle BE, DF is equal to the parallelogram AC. 14. A, B, C, D are four points in the same straight line, E a point in that line equally distant from the middle of the segments AB, CD; F is any other whatever in AD; then AF2 + BF2 + CF2 + DF2 AE2+ BE2+ CE2 + DE2 + 4 EF2. = 15. The sum of the perpendiculars let fall from any points within an equilateral triangle, will be equal to the perpendicular let fall from one of its angles upon the opposite side. Is this proposition true when the point is in one of the sides of the triangle? In what manner must the proposition be enunciated when the point is without the triangle? 16. If a line AB be divided into two parts AC and CB in the point C (Prop. 11, Book 11.), so that the rectangle AB × BC = AC2 ; and if AC be divided in D, so that CD=BC, prove that AC× AD=BC3. 17. All plane rectilineal figures admit of quadrature. Point out the succession of steps by which Euclid establishes the truth of this proposition. 18. The hypothenuse (c) 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 of the sides of the triangle CDE = ? .c2. 19. If from the right angle C of a right-angled triangle ABC, straight lines be drawn to the opposite angles of the square described on the hypothenuse AB; shew that the difference of the squares described on these lines is equal to the difference of the squares described on the two sides AC, BC. 20. If the sides of the triangle be as the numbers 2, 4, 5, shew whether it will be acute or obtuse-angled. 21. If an angle of a triangle be two-thirds of two right angles, shew that the square of the side subtending that angle is equal to the squares of the sides containing it, together with the rectangle contained by those sides. 22. In any triangle the squares of the two sides are together double of the two squares of half the base and of the straight line joining its bisection with the opposite angle. 23. The square described on a straight line drawn from one of the angles at the base of a triangle to the middle point of the opposite side, is equal to the sum or difference of the square of half the side bisected and the rectangle contained between the base and that part of it, or of it produced, which is intercepted between the same angle and a perpendicular drawn from the vertex. 24. If perpendiculars be drawn from the extremities of the base of a triangle on a straight line which bisects the angle opposite to the base, the area of the triangle is equal to the rectangle contained by either of the perpendiculars and the segment of the bisecting line between the angle and the other perpendicular. 25. Upon the sides AB, BC, CA of the triangle ABC, or upon these produced, let fall the perpendiculars DE, DF, DG, from the point D within or without the triangle. Then AE2+ BF2+ GC2 = BE+CF+ AG. Required a demonstration. 26. If from the three angles of a triangle, lines be drawn to the points of bisection of the opposite sides, the squares of the distances between the angles and the common intersection are together one-third of the squares of the sides of the triangle. 27. Prove that the square of any straight line drawn from the vertex of an isosceles triangle to the base, is less than the square of a side of the triangle by the rectangle contained by the segments of the base. 28. If from one of the equal angles of an isosceles triangle a perpendicular be drawn to the opposite side, the rectangle contained by that side and the segment of it intercepted between the perpendicular and base is equal to half the square described upon the base. 29. If in an isosceles triangle a perpendicular be let fall from one of the equal angles to the opposite side, the square of the perpendicular is equal to the square of the line intercepted between the other equal angle and the perpendicular, together with twice the rectangle contained by the segments of that side. 30. The square on the base of an isosceles triangle whose ver tical angle is a right angle, is equal to four times the area of the triangle. == 31. If ABC be an isosceles triangle, and CD be drawn perpendicular to AB; the sum of the squares of the three sides = AD2+ 2.′BD3 + 3.CD3. 32. If ABC be an isosceles triangle, and DE be drawn parallel to the base BC, and EB joined; prove that BE2= BC × DE + CE3. 33. If ABC be an isosceles triangle of which the angles at B and Care each double of A; then the square of AC is equal to the square of BC together with the rectangle contained by AC and BC. 34. Shew that in a parallelogram the squares of the diagonals are equal to the sum of the squares of all the sides. 35. If ABCD be any rectangle, A and C being opposite angles, and O any point either within or without the rectangle: OA2 + OC2 = OB2 + OD2. 36. In any quadrilateral figure, the sum of the squares of the diagonals together with four times the square of the line joining their middle points is equal to the sum of the squares of all the sides. 37. In any trapezium, if the opposite sides be bisected, the sum of the squares of the two other sides, together with the squares of the diagonals, is equal to the sum of the squares of the bisected sides together with four times the square of the line, joining the points of bisection. 38. The squares of the diagonals of a trapezium are together double the squares of the two lines joining the bisections of the opposite sides. 39. In any trapezium two of whose sides are parallel, the squares of the diagonals are together equal to the squares of its two sides which are not parallel, and twice the rectangle contained by the sides which are parallel. 40. If squares be described on the sides of any triangle and the angular points of the squares be joined; the sum of the squares of the sides of the hexagonal figure thus formed is equal to four times the sum of the squares of the sides of the triangle. |