5. The square on the sum of two straight lines exceeds the square on their difference by four times the rectangle contained by the two lines. 6. The square on the sum of two straight lines, together with the square on their difference, is equal to twice the sum of the squares on the two straight lines. Use Euc. II. 4 to show that (AB+CD=AB2+CD2+2AB.CD, and Ex. 2 or Euc. II. 7 to show that (AB-CD=AB2+CD2-2AB.CD, and add the results. For a geometrical proof see $ 35, Ex. 2. 7. If the straight line AB be bisected in C and produced to D, so that the square on BD is equal to twice the square on BC, show that the rectangle AD.BD shall be equal to the rectangle AB-CD. 8. AB is divided in P and Q, so that AP=QB; show that AB2+PQ2=2AP2+2PB2. 9. ABCD is a quadrilateral, right-angled at A and C. From B and D, BM and DN are drawn AC. Prove AM=CN. Use Euc. I. 47, § 20, Ex. 5, and § 33, Ex. 1. 10. A, B, C, D are four points in order in the same straight line, and AB=2BD; show that AC2+2CD2=AB2+2BD2+3BC2. $ 35, (Bookwork, Euclid II. 1-10.) 1. ABC is an isosceles right-angled triangle, and P is any point in the hypotenuse BC, or in BC produced ; show that BP2+PC=2 AP2. D is the mid-point of the hypotenuse of the right-angled A ABC. .: AD=BD=CD. [S 12. Ex. 1. and since the A ABC is isosceles, the median ADI BC. [$ 3, Ex. 1. In either figure, = 2AD2+2DP2 [Euc. II. 9 and 10. [Proved above. [Euc. I. 47. 2. The square on the sum of two straight lines, together with the square on their difference, is equal to twice the sum of the squares on the two straight lines. (A Standard Theorem.) Show that Euclid II. 9 and 10 are particular cases of this general theorem, and prove it by a method which shows how the one set of squares can be fitted together so as to form the other set. Let AC, CD be the two straight lines, produce CD to E so that CE=AC, and cut off AB=DE, then we have to show that (AC+CD)2+(AC-CD)2=2 AC2+2 CD2; that is, that AD2+DE2=2 AC2+2 CD2 Since AB=DE, this corresponds with II. 9, if we consider AE as the line bisected in C and divided in D, while it corresponds with II. 10, if we consider BD as the line bisected in C and divided externally in A. In figure 1, describe squares on AD, DE, AC, CE, KL, and prove FK=MP, and hence Al+DP=AL+ON+KH+LI; etc. Otherwise In figure 2, describe squares on AD, DE, AC, CD, FJ, and prove RL=LS=DE, and hence that the gnomon QT+DP=AL, etc. This latter construction is due to Miss Hilda Hudson. See Nature, December 24, 1891. Observe, also, that since AC={(AD+DE), CD={(AD-DE), the theorem may also be enunciated thus : The sum of the squares on two straight lines is equal to twice the square on half the sum of the straight lines, together with twice the square on half their difference. 3. Points H and K are taken in the base BC of an isosceles right-angled A ABC, such that BK=CH=AB; show that (1) HK=BHP+KC?; (2) BC?=BH2+KC2+2BC.HK. (1) Use Euc. II. 9, and show that BH2= 2HD?, etc. (2) Use Euc. II. 4, etc. D 4. A, B, C, D are four points in order in the same straight line, and AB=BD; show that AC2+CD2=4 BC2+2AC CD. 5. A, B, C, D are four points in order in the same straight line, and AB=BC; show that AD2+CD2=4 BD2 - 2AD.CD. Compare with preceding Ex. in view of the notes to $ 34, Exx. 1, 2. 6. A, B, C, D are four points in order in the same straight line, and AB>CD; show that AC2+CD2>AB2+BD2. 7. ABCD is a square. If E be a point in AC, or AC produced, show that the triangle whose sides are AE, CE, and the diagonal of the square on BE is right-angled. Prove that AE2+ EC2=2BE?. 8. ABC is an isosceles triangle, and P is any point in the base BC, or BC produced; show that BP2+CP2+2AB2 - 2AP2=BC2. $36. (Bookwork, EUCLID II., 1-11.) 1. If, in the figure of Euclid II. 11, AD be produced to F, so that EF=EB, and BA be produced to H' so that H'A=AF', then AH'2=AB H'B. GP = Complete the square AH'G'F', and produce CD to meet H'G' in K'. AF.DF+AE2=EF2; [Euc. II. 6. =EB2; [Hypothesis. =AE2+AB?; [Euc. I. 47. .: AF'.DF=AB2; [Subtracting equals. or DF'G'K'=ABCD. ... AFG'H'=BCK'H'; or AH'?=BC.H'B=AB.H'B. DEFINITION.—When a straight line is divided, either internally or externally, into two segments, such that the rectangle contained by the whole line and one segment is equal to the square on the other segment, it is said to be divided in Medial Section. In the above figure AH2=AB.HB, [Euc. II. 11. and AH'2=AB.H'B. [Proved above. Thus AB is divided internally at H and externally at H' in medial section. |