15. Let ABC be a triangle of the system on the fixed base AB. Produce AC to D, making CD equal to CB. Then AD is of constant magnitude. Join BD cutting the bisector of the BCD at P. Then CP bisects BD at rt. angles. Required the locus of P. Bisect AB at O. Join OP. Then OP is one-half of AD [Ex. 3, p. 97]. That is, OP is constant; and since O is a fixed point, the locus of P is a circle, whose centre is at O, and whose radius is half AD. 16. Join AQ, and produce it to meet A'P' at X. Then, by hyp. and III. 31, AX, PP', A'Q' are par1. Also QQ and A'X are par1. Then 17. Let X, Y ce of the C (X). AA'2 = AX2 + XA' [1. 47] = P'P2 + Q'Q2. Join A'Q'. be the centres of the two O3, P being on the Join AB, AX, AY, AD. Then AX and AY are tangents [Hyp.]. Hence, by addition, ▲ XAY = ▲ ADC + ACD. But XAY is a rt. angle [Hyp.]; hence DAC is a rt. angle [1. 32]; .. DC is a diameter [III. 31]. 18. Join PA, PB, PC; and bisect these lines at S1, S2, S. Then since the s P, F, E, A lie on a of the PEA, PFA are rt. angles, .. the four points whose diam. is PA. about the AEPF. Similarly for S2 and S.. Hence S, is the centre Again [Ex. 2 and 3, p. 96], S,S, SS, SS, are par1. to AB, BC, CA, and equal to half of these lines, .. the ▲ SSS, is equiangular to the ABC. 19. Take O the centre of the O. Join PC: then PC pro duced must pass through O. Join OA, OX, OY. S Then since the $ PAO, ACO are rt. angles, .. PA must touch the circumscribed about the ▲ ACO. .. PC. PO = PAPX. PY [III. 36]. ..the four points X, C, O, Y are concyclic. And AC is perp. to PO; .. CA bisects the XCY. 20. Let AB be the sum of the lines, K the side of the sq. to which the rectangle is equal. On AB describe a semi-, and draw a st. line par1. to AB at a distance from it equal to K, cutting the semi- at P, P'. From P (or P') draw PX perp. to AB. Then AX, XB are the required st. lines; for AX. XB = PX2. This may be proved as in II. 14, or as a special case of III. 35. 21. Let the sum of the sqq. on required lines be equal to the sq. on AB, and the rectangle contained by them to the sq. on K. Analysis. On AB describe a semi-: then if any point P is taken on the ◇ce, we shall have AP2 + PB2 = AB2 [111. 31, 1. 47]. Hence we have to find a point P on the ce such that AP. PB = sq. on K. Suppose PX drawn perp. to AB. Then ▲ APB = |rect. AP, PB; for APB is a rt. angle. Also AAPBrect. AB, PX [1. 41]. Hence rect. AP, PB = rect. AB, PX. Construction. [1.45]. To AB apply a rectangle equal to the sq. on K And let D be its altitude. Draw MN par1. to AB at a distance from it equal to D, cutting the semi- at P or P'. Then evidently AP, BP are the lines required. 22. Let K be the sum of the required lines, and let the sq. on AB be equal to the sum of the sqq. on them. On AB describe a semi-circle, and also a segment containing an angle equal to half a rt. angle. From centre A, with radius K, draw a cutting the latter segment at D (or D'). Join AD cutting the semi- at P. Join PB. Then shall AP, BP be the required lines. Join DB. 23. Let AB be the diff. of the required lines, and let the rect. contained by them be equal to the sq. on K. On AB as diam. describe a O. ce At any point T on the Oce draw a tangent TP, making TP equal to K. Take the centre O, and draw PQOR cutting the O at P, Q. Then shall PQ, PR be the required lines. For rect. PR, PQ = the sq. on PT [III, 36] And the diff. of PR and PQ is QR, that is, AB. 24. Let AB be the sum or diff. of the required lines, and the sq. on CD the diff. of the sqq. on them. Draw DE perp. to CD, and of any length. Join CE. Then CD2 = CE2 - ED2 [1. 47]. From centre A, with radius CE, describe a O. From centre B, with radius DE, describe a O, cutting the other at P (or P'). From P draw PX perp. to AB, or AB produced. Then AX and BX shall be the required lines. For AX2 + PX2 = AP3, and BX2 + PX2 = BP2 [1. 47]; -- And 25. The So that AX + BX = AB. PAP' must be a rt. angle [Ex. 2, p. 29]. OAC = 4 OAP-L CAP =OPA-PAB [1. 5 and Hyp.] = LABC [1. 32], .. OA is a tangent to the C about the AABC [III. 32]. 26. Let the feet of the perps. be D, E, F. [Take the figure, as on p. 232, in which F is in AB produced.] (i) The four points E, C, P, D are evidently concyclic, That is, .. LECD = EPD [III. 21]. ▲ ACB = LA'PB'. Hence arc A'B' arc AB. Hence chord A'B' = chord AB. = And so for the other sides. Hence the As are identically equal. (ii) Join A'B. Since arc AB = arc A'B', .. AA'B = LA'BB' [III. 27]; .. AA' is par1. to BB' [1. 27]. 27. Take two lines of the system PQ and pq. It may be proved by method similar to that of Ex. 1, p. 196, that arc Pparc Qq. Hence the chord Pp is equal and par1. to chord Qq. From which it follows that PQ=pq [1. 33]. The prop. may also be proved by noting that P is the orthocentre of the AQB [Ex. 35 and Ex. 31, p. 227]. 28. With figure of p. 225, let S be the centre of circum. ©, and let SA meet EF at X. 8 Hence from AFX, the AXF is a rt. angle [1. 32]. H. K. E. 29. Since the CEP, CDP are rt. angles, the about the APED passes through C, and is described on PC as diam. Hence it is required to find the locus of X, the middle point of CP. Take S the centre of the O, and join SX. Then since the CXS is a rt. angle [III. 3], and the points C, X are fixed, .. the locus is a on CS as diam. 30. join AP, PX. Take the figure of p. 232. Draw the diam. AX, and Then the four points P, D, E, C are concyclic. 31. Let ACD, AEF and BEC, BFD be the two pairs of lines. Let the about the As ACE, BEF meet at P. 8 Then shall the Os about the As AFD, BCD pass through P. Join PF, PE, PA. Hence S = PAD, PFD together two rt. angles. .. the points A, D, F, P are concyclic; that is, the about the ADF passes through P. The proposition may also be proved by the properties of Simson's Line [See Exx. 74, 77, p. 232]. 8 32. From the last exercise it is seen that the Os about the four ▲ pass through a common point P. Hence it may be seen (Ex. 77, p. 232) that the four ▲3 have a common pedal for the point P. Also (Ex. 78, p. 233) this pedal bisects each of the lines joining P to the four orthocentres. Hence, by the method of Ex. 2, p. 116, the orthocentres are collinear. |