Obs. If the perpendicular AD coincides with AB, that is, if ABC is a right angle, then twice the rect. BC, CD becomes twice the sq. on BC; and it may be shewn that the proposition merely repeats the result of 1. 47. NOTES ON PROP. 13. (i) Remembering the definition of the projection of a straight line given on p. 153, we may enunciate Prop. 13 as follows; In every triangle, the square on the side subtending an acute angle is less than the sum of the squares on the sides containing that angle, by twice the rectangle contained by one of these sides and the projection of the other side upon it. (ii) Comparing the Enunciations of 11. 12, 1. 47, 11. 13, we see that in the triangle ABC, if the angle ACB is obtuse, we have by II. 12, AB2= BC2 + CAP + 2 BC. CD; AB2= BC2 + CAP; AB2= BC2 + CA2 – 2 BC. CD. The square on a side of a triangle is greater than, equal to, or less than the sum of the squares on the other sides, according as the angle opposite to the first is obtuse, a right angle, or acute. EXERCISES ON II. 12 AND 13. 1. If from one of the base angles of an isosceles triangle a perpendicular is drawn to the opposite side, then twice the rectangle contained by that side and the segment adjacent to the base is equal to the square on the base. 2. If one angle of a triangle is one-third of two right angles, shew that the square on the opposite side is less than the sum of the squares on the sides forming that angle, by the rectangle contained by these two sides. [See Ex. 10, p. 109.] 3. If one angle of a triangle is two-thirds of two right angles, shew that the square on the opposite side is greater than the sum of the squares on the sides forming that angle, by the rectangle contained by these sides. [See Ex. 10, p. 109.] PROPOSITION 14. PROBLEM. To describe a square that shall be equal to a given rectilineal figure. H Let A be the given rectilineal figure. It is required to describe a square equal to A. Construction. Describe a parTM BCDE equal to the fig. A, and having the angle CBE a right angle. I. 45. Then if BC=BE, the fig. BD is a square; and what was required is done. But if not, produce BE to F, making EF equal to ED; I. 3. and bisect BF at G. 1. 10. With centre G, and radius GF, describe the semicircle BHF; produce De to meet the semicircle at H. Then shall the sq. on EH be equal to the given fig. A. Join GH. Proof. Because BF is divided equally at G and unequally at E, .. the rect. BE, EF with the sq. on GE= the sq. on GF II. 5. the sq. on GH. But the sq. on GH = the sum of the sqq. on GE, EH; for the angle HEG is a rt. angle. ... the rect. BE, EF with the sq. on GE= the sum of the sqq. on GE, EH. From these equals take the sq. on GE : then the rect. BE, EF=the sq. on HE. But the rect. BE, EF=the fig. BD; for EF=ED; Constr. and the fig. BD= the given fig. A. Constr. .. the sq. on EH = the given fig. A. Q.E.F. I. 47. QUESTIONS FOR REVISION ON BOOK II. 1. Explain the phrase, the rectangle contained by AB, CD; and shew by superposition that if AB=PQ, and CD=RS, then the rectangle contained by AB, CD=the rectangle contained by PQ, RS. 2. Shew that Prop. 2 is a special case of Prop. 1, explaining under what conditions Prop. 1 becomes identical with Prop. 2. 3. What must be the relation between the divided and undivided lines in the enunciation of Prop. 1 in order to give the result proved in Prop. 3? 4. Define the segments into which a straight line is divided at a point in such a way as to be applicable to the case when the dividing point is in the given line produced. Hence frame a statement which includes the enunciations of both II. 5 and 11. 6, and find the algebraical formulae corresponding to these enunciations. Also combine in a single enunciation the results of 1. 9 and 11. 10. 5. Compare the results proved in Propositions 4 and 7 by finding the algebraical formulae corresponding to their enunciations. 6. The difference of the squares on two straight lines is equal to the rectangle contained by their sum and difference. Deduce this theorem from Prop. 5. 7. Define the projection of one straight line on another. How may the enunciations of 11. 12 and 11. 13 be simplified by means of this definition ? 8. In the figure of Proposition 14, (i) If BE=8 inches, and ED=2 inches, find the length of EH. (ii) If BE=12:5 inches, and EH=2.5 inches, find the length of ED. (iii) If BE=9 inches, and EH=3 inches, find the length of GH. 9. When is a straight line said to be divided in medial section? If a straight line 8 inches in length is divided internally in medial section, shew that the lengths of the segments are approximately 4.9 inches and 3•1 inches. [Frame a quadratic equation as explained on page 151, and solve.] THEOREMS AND EXAMPLES ON BOOK II. ON II. 4 AND 7. 1. Shew by II. 4 that the square on a straight line is four times the square on half the line. [This result is constantly used in solving examples on Book II., especially those which follow from 11. 12 and 13.] 2. If a straight line is divided into any three parts, the square on the whole line is equal to the sum of the squares on the three parts together with twice the rectangles contained by each pair of these parts. Shew that the algebraical formula corresponding to this theorem is (a+b+c)2=a? +62+c2 + 2bc+2ca + 2ab. 3. In a right-angled triangle, if a perpendicular is drawn from the right angle to the hypotenuse, the square on this perpendicular is equal to the rectangle contained by the segments of the hypotenuse. 4. In an isosceles triangle, if a perpendicular is drawn from one of the angles at the base to the opposite side, shew that the square on the perpendicular is equal to twice the rectangle contained by the segments of that side together with the square on the segment adjacent to the base. 5. Any rectangle is half the rectangle contained by the diagonals of the squares described upon its two sides. 6. In any triangle if a perpendicular is drawn from the vertical angle to the base, the sum of the squares on the sides forming that angle, together with twice the rectangle contained by the segments of the base, is equal to the square on the base together with twice the square on the perpendicular. ON II. 5 AND 6. Obs. The student is reminded that these important propositions are both included in the following enunciation : The difference of the squares on two straight lines is equal to the rectangle contained by their sum and difference. [See Cor., 7. In a right-angled triangle the square on one of the sides forming the right angle is equal to the rectangle contained by the sum and difference of the hypotenuse and the other side. [1. 47 and 11. 5, Cor.] p. 137]. 8. The difference of the squares on two sides of a triangle is equal to twice the rectangle contained by the base and the intercept between the middle point of the base and the foot of the perpendicular drawn from the vertical angle to the base. Let ABC be a triangle, and let P be the middle point of the base BC: let AQ be drawn perp. to BC. Then shall AB2 – AC2=2BC. PQ. I. 47. Ax. 3. =(BQ+QC)(BQ - QC) 11. 5, Cor. =BC. 2PQ Ex., p. 137. =2BC. PQ Q.E.D. The case in which AQ falls outside the triangle presents no difficulty. 9. The square on any straight line drawn from the vertex of an isosceles triangle to the base is less than the square on one of the equal sides by the rectangle contained by the segments of the base. 10. The square on any straight line drawn from the vertex of an isosceles triangle to the base produced, is greater than the square on one of the equal sides by the rectangle contained by the segments into which the base is divided externally. 11. If a straight line is drawn through one of the angles of an equilateral triangle to meet the opposite side produced, so that the rectangle contained by the segments of the base is equal to the square on the side of the triangle ; shew that the square on the line so drawn is double of the square on a side of the triangle. 12. If XY is drawn parallel to the base BC of an isosceles triangle ABC, then the difference of the squares on BY and CY is equal to the rectangle contained by BC, XY. [See above, Ex. 8.] 13. In a right-angled triangle, if a perpendicular is drawn from the right angle to the hypotenuse, the square on either side forming the right angle is equal to the rectangle contained by the hypotenuse and the segment of it adjacent to that side. |