PROPOSITION I. PROBLEM. Divide a given straight line into two parts such, that their rectangle may be equal to a given square; and determine the greatest square which the rectangle can equal. Let AB be the given straight line, and let M be the side of the given square. It is required to divide the line AB into two parts, so that the rectangle contained by them may be equal to the square on M. Bisect AB in C, with center C, and radius CA or CB, describe the semicircle ADB. At the point B draw BE at right angles to AB and equal to M. Through E, draw ED parallel to AB and cutting the semicircle in D; and draw DF parallel to EB meeting AB in F. Then AB is divided in F, so that the rectangle AF, FB is equal to the square on M. (II. 14.) The square will be the greatest, when ED touches the semicircle, or when M is equal to half of the given line AB. PROPOSITION II. THEOREM. The square on the excess of one straight line above another is less than the squares on the two lines by twice their rectangle. Let AB, BC be the two straight lines, whose difference is AC. Then the square on AC is less than the squares on AB and BC by twice the rectangle contained by AB and BC. Constructing as in Prop. 4. Book II. Because the complement AG is equal to GE, add to each CK, therefore the whole AK is equal to the whole CE; (a + x) (α is equivalent to t to the two prop 35. Prove B on AC pr to the recta 36. Ir 37. the obt is equ pro 38 T ible of AK; he difference ngle AB, BC. LON III. THEOREM. quares on the two sides are together double of the use and on the straight line joining its bisection with the Cbe a triangle, and AD the line drawn from the vertex A Disection D of the base BC. A ... four t squa an there tim a B D E C From A draw AE perpendicular to BC. Then, in the obtuse-angled triangle ABD, (II. 12.); the square on AB exceeds the squares on AD, DB, by twice the and in the acute-angled triangle ADC, (II. 13.); the square on AC is less than the squares on AD, DC, by twice wherefore, since the rectangle BD, DE is equal to the rectangle CD, PROPOSITION IV. THEOREM. If straight lines be drawn from each angle of a triangle bisecting 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. Let ABC be any triangle, and let AD, BE, CF be drawn from A, B, C, to D, E, F, the bisections of the opposite sides of the triangle: draw AG perpendicular to BC. T teral ang one enu 117 quare c rect 'ar 'ifference of the rectangles of the sides parallelogram, of which A, C are Fany point in CD. Prove that together with the rectangle BE, e squ Similarly, twice the squares of square on FC: hence, by adding these equ four times the squares on AB, AC, BC are equ squares on AD, BE, CFtogether with the squares and taking the squares on AB, AC, BC from thes e joined cutting therefore three times the squares on AB, AC, BC are eq times the squares on AD, BE, CF. vn to the istances >-third PROPOSITION V. THEOREM. The sum of the perpendiculars let fall from any point within an equila the E one of its teral triangle, will be equal to the perpendicular let fall from 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? 9 Let ABC be an equilateral triangle, and P any point within it: and from Plet fall PD, PE, PF perpendiculars on the sides AB, BC, CA respectively, also from A let fall AG perpendicular on the base BC. Then AG is equal to the sum of PD, PE, PF. From P draw PA, PB, PC to the angles A, B, C. Then the triangle ABC is equal to the three triangles PAB, PBC, РСА. and AK, CE together are double of AK; therefore the difference of AE and the gnomon AKF is equal to the difference between the squares on AB, BC, and twice the rectangle AB, BC; but the difference AE and the gnomon AKF is the figure HF, which is equal to the square on AC. Wherefore the square on AC is equal to the difference between the squares on AB, BC, and twice the rectangle AB, BC. PROPOSITION III. THEOREM. In any triangle the squares on the two sides are together double of the squares on half the base and on the straight line joining its bisection with the opposite angle. Let ABC be a triangle, and AD the line drawn from the vertex A to the bisection D of the base BC. From A draw AE perpendicular to BC. Then, in the obtuse-angled triangle ABD, (II. 12.); the square on AB exceeds the squares on AD, DB, by twice the rectangle BD, DE: and in the acute-angled triangle ADC, (II. 13.); the square on AC is less than the squares on AD, DC, by twice the rectangle CD, DE: wherefore, since the rectangle BD, DE is equal to the rectangle CD, DE; it follows that the squares on AB, AC are double of the squares on AD, DB. PROPOSITION IV. THEOREM. If straight lines be drawn from each angle of a triangle bisecting 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. Let ABC be any triangle, and let AD, BE, CF be drawn from A, B, C, to D, E, F, the bisections of the opposite sides of the triangle: draw AG perpendicular to BC. Then the square on AB is equal to the squares on BD, DA together with twice the rectangle BD, DG, (II. 12.) and the square on AC is equal to the squares on CD, DA diminished by twice the rectangle CD, DĞ; (II. 13.) therefore the squares on AB, AC are equal to twice the square on BD, and twice the square on AD; for DC is equal to BD: and twice the squares on AB, AC are equal to the square on BC, and four times the square on AD: for BC is twice BD. Similarly, twice the squares on AB, BC are equal to the square on AC, and four times the square on BE: also twice the squares on BC, CA are equal to the square on AB, and four times the square on FC: hence, by adding these equals, four times the squares on AB, AC, BC are equal to four times the squares on AD, BE, CFtogether with the squares on AB, AC, BC: and taking the squares on AB, AC, BC from these equals, therefore three times the squares on AB, AC, BC are equal to four times the squares on AD, BE, CF. PROPOSITION V. THEOREM. The sum of the perpendiculars let fall from any point 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? Let ABC be an equilateral triangle, and P any point within it: and from Plet fall PD, PE, PF perpendiculars on the sides AB, BC, CA respectively, also from A let fall AG perpendicular on the base BC. Then AG is equal to the sum of PD, PE, PF. From P draw PA, PB, PC to the angles A, B, C. Then the triangle ABC is equal to the three triangles PAB, PBC, РСА. |