THEOREM 53. [Euclid II. 5 and 6.] The difference of the squares on two straight lines is equal to the rectangle contained by their sum and difference. Let the given lines AB, AC be placed in the same st. line, and let them contain a and b units of length respectively, It is required to prove that namely that Construction. On AB and AC draw the ACFG; and produce CF to meet ED at H. Then GE CB = a - b units. = squares ABDE, Proof. Now AB2 - AC2 = the sq. AD - the sq. AF COROLLARY. If a straight line is bisected, and also divided (internally or externally) into two unequal segments, the rectangle contained by these segments is equal to the difference of the squares on half the line and on the line between the points of section. That is, if AB is bisected at X and also divided at Y, internally in Fig. 1, and externally in Fig. 2, then 1. Draw diagrams on squared paper to shew that the square on a straight line is X 2. (i) four-times the square on half the line; (ii) nine-times the square on one-third of the line. Draw diagrams on squared paper to illustrate the following algebraical formula: (i) (x+7)2= x2+14x+49. (ii) (a+b+c)2=a2+b2+c2+2bc+2ca+2ab. (iii) (a+b)(c+d) = ac+ad+bc+bd. (iv) (x+7)(x+9) = x2 + 16x + 63. 3. In Theor. 50, Cor. (i), if AB=4 cm., and the fig. AE=96 sq. cm., find the area of the fig. XC. X 4. In Theor. 50, Cor. (ii), if AX=2∙I", and the fig. XC=3:36 sq. in., find AB. 5. In Theor. 51, if the fig. AG=36 sq. cm., and the rect. AX, XB =24 sq. cm., find AB. 6. In Theorem 52, if the fig. AG=9.61 sq. in., and the fig. DG=6.51 sq. in., find AB. H.S G. [For further Examples on Theorems 50-53 see p. 230.] P THEOREM 54. [Euclid II. 12.] In an obtuse-angled triangle, the square on the side subtending the obtuse angle is equal to the sum of the squares on the sides containing the obtuse angle together with twice the rectangle contained by one of those sides and the projection of the other side upon it. A B D project. on B Let ABC be a triangle obtuse-angled at C; and let AD be drawn perp. to BC produced, so that CD is the projection of the side CA on BC. [See Def. p. 63.] It is required to prove that AB2 = BC2 + CA2 + 2BC. CD. Proof. Because BD is the sum of the lines BC, CD, .. BD2 = BC2 + CD2 + 2BC. CD. To each of these equals add DA2. Then BD2 + DA2 = BC2 + (CD2 + DA2) + 2BC. CD. Theor. 51. THEOREM 55. [Euclid II. 13.] In every triangle the square on the side subtending an acute angle is equal to the sum of the squares on the sides containing that angle diminished by twice the rectangle contained by one of those sides and the projection of the other side upon it. AC projected on to Let ABC be a triangle in which the C is acute; and let AD be drawn perp. to BC, or BC produced; so that CD is the projection of the side CA on BC. Proof. Since in both figures BD is the difference of the lines BC, CD, ... BD2 = BC2 + CD2 – 2BC. CD. To each of these equals add DA2. Theor. 52. Then BD2 + DA2 = BC2 + (CD2 + DA2) - 2BC. CD. ....(i) Q.E.D SUMMARY OF THEOREMS 29, 54 and 55. 444 (i) If the ACB is obtuse, C(D) B Observe that in (ii), when the LACB is right, AD coincides with AC, so that CD (the projection of CA) vanishes ; hence, in this case, 2BC. CD = 0. Thus the three results may be collected in a single enunciation: 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 contained by those sides is obtuse, a right angle, or acute; the difference in cases of inequality being twice the rectangle contained by one of the two sides and the projection on it of the other. EXERCISES. 1. In a triangle ABC, a=21 cm., b=17 cm., c= 10 cm. By how many square centimetres does c2 fall short of a2+b2? Hence or otherwise calculate the projection of AC on BC. 2. ABC is an isosceles triangle in which AB=AC; and BE is drawn perpendicular to AC. Shew that BC2=2AC. CE. 3. In the ABC, shew that (i) if the LC-60°, then c2=a2+b2 - ab; |