4. Let the ABC have the ABC equal to two-thirds of two rt. angles. Draw AX perp. to CB produced. Then the ABX = one-third of two rt. angles; hence, as in the last Ex., AB = 2BX. And by II. 12, AC2 AB2 + BC2 + 2BX.BC = AB2+ BC2 + AB. BC. THEOREMS AND EXAMPLES ON BOOK II. ON II. 4 AND 7. 1. Let AB be the st. line, and C its middle point, then by 11. 4 AB2 = AC2 + CB2 + 2AC. CB 2. Let AB be divided into three parts at the points X, Y. On AB describe the sq. ABCD. Join BD. Through X and Y draw XP, YQ par1. to AD or BC and cutting BD at H and K. Through H and K draw LHMN and EFKG par1. to AB. Then prove as in II. 4 that (i) the figs. LP, FM, YG are respectively the sqq. on AX, XY, YB. (ii) the fig. AF the fig. MC = the rect. AX, YB. (iii) the fig. EH = the fig. HQ: = (iv) the fig. XK = the fig. KN = the rect. AX, XY. the rect. XY, YB. 3. Let ABC be a ▲ rt. angled at B, and let BD be drawn perp. to AC. 4. Let ABC be an isosceles A, having its vertex at A. Draw BD perp. to AC. 5. Let ABCD be a rectangle; and on AB, BC let sqq. ABGH, BCEF be described externally to the rectangle. Join HB, BE. Then HB, BE are clearly in one line, for the < ABH, CBE are each half of a rt. angle. 8 So that But, as in II. 9, or, HB2+ BE2 + 2HB. BE = 2HA2 + 2AD2 + 4HA. AD. HB2 = 2HA2, and BE2 = 2BC2 = 2AD2, .. 2HB. BE = 4HA. AD ; the rect. HB. BE = twice the rectangle ABCD. 6. Let ABC be the given A. Then by II. 4, From A draw AX perp. to BC. BC2 = BX2 + XC2 + 2BX. XC. To each of these equals add 2 . AX2. So that BC2 + 2AX2 = BX2+ AX2 + XC2 + AX2 + 2BX. XC = AB2 + AC2 + 2BX. XC. [1. 47.] 9. Let ABC be an isosceles ▲, having its vertex at A; and let AQ be drawn to any point Q in BC. or, Draw AP perp. to BC. Then, by 1. 26, BP PC. Hence by II. 5, rect. BQ, QC + PQ2 = PC2. To each of these equals, add AP2; so that BQ. QC + PQ2 + AP2 = PC2 + AP2; BQ. QC + AQ2 = AC2. [1. 47.] 10. Let ABC be the isosceles triangle having its vertex at A, and let Q be any point in the base BC produced. Join AQ, and draw AP perp. to BC. Then it may be shewn by 1. 26 that BC is bisected at P. Hence by II. 6, BQ. QC + PC2 = PQ2. To each of these equals add AP2. .`. BQ . QC + PC2 + AP2 = PQ2 + AP2 ; or, BQ. QC + AC2 = AQ2. that 11. Taking the same letters as in the last Ex. we may or, or, But AC2 + CX2 = 2AC. CX + AX2. AC2 = AB2 + BC2 2BX2 + AX2 + XC2 [1. 47]. .. 2BX2 + 2CX2 + AX2 = 2AC. CX + AX2; BX2 + CX2 = AC. CX ; BC2 = AC. CX. [1. 47.] shew ON II. 9 AND 10. 14. Let AB be the given st. line divided equally at P and unequally at Q. Then AB2 = 4AP2; Hence also AB2 = AQ2 + QB2 + 2AQ. QB. [11. 4.] AQ2 + QB2 = 4AP2 – 2AQ. QB. or, But AQ. QB = AP2 - PQ2 [11. 5]. .. AQ2 + QB2 = 4AP2 - 2AP2 + 2PQ2; 15. Let the st. Then, as before, also by II. 7, But by II. 6, line AB be bisected at P and produced to Q. AB2 = 4AP2; AQ2 + QB2 = 2AQ. QB + AB2. .. AQ2 + QB2 = 2AQ. QB + 4AP2. AQ. QB = PQ2 — AP2. .', AQ2 + QB2 = 2PQ2 - 2AP2 + 4AP2 16. Let the given st. line AB be divided equally at P and unequally at Q. 17. Let AB be divided in medial section at H. From A cut off AH' equal to BH. That is, AH is divided in medial section at H'. then 18. If AB is divided in medial section at H, = AH. HB + HB2 - HB2 [11. 3] = But = .. AH = XB – XH, AH2 + XH2 = XB2. .. the ▲, whose sides are AH, XH and XB, is right-angled. or, By addition AB2 + AC2 = AB2 + AC2 + 2BC2 - 2AB. BF - 2AC.CE, |