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25. Let o be the given pt., AB, CD the par!. lines. With A as centre and radius equal to given length describe a circle. This will in general cut cd in two pts. L, M. Then lines drawn through o par. to AL and AM will be the required lines.
26. Let AB be the line to which the required line is to be par!., QP and RS the other two given lines. Let QP meet AB in P; draw PT on AB equal to the given length; through T draw TR par!. to PQ meeting SR in R; through R draw RQ par! to AB meeting PQ in Q. Then QR is the required line (1. 33].
27. Let the given lines PO, QO meet in O; bisect POQ by OS. Draw OR perp. to os on the same side as OQ and equal to the given length; through R draw RT pard. to op meeting oQ in T. Through T draw TV parl. to OR meeting OP in V. Then TV is equal and par. to OR [1. 33), and since it is perp. to os it is equally inclined to OP and OQ (1. 26).
28. Join AP; bisect AP in 2, and draw QR par!. to AB (the further line) meeting AC in R. Join PR and produce it to meet AB in S. Then QR bisects PS [Ex. 1, p. 96].
1. – ACD = L ADC (1. 5] and L ACB - ABC; ... BCD=sum of 48 CBD, BDC. That is, Z BCD is a rt. — [1. 32].
2. Let CD join the rt. Lc to D the middle pt. of AB. Draw DE parl. to AC meetin BC in E. Then BC is bisected at E [Ex. 1, p. 96). Also _DEB, DEC are equal, being rt. 29. :: A' DEB, DEC are identically equal [1. 4].
.. DC = DB. 3. By Ex. 2 each of the lines is equal to half the base. 4. By Ex. 2, p. 96, AZYX is a parm. . . ZXY = L BAC. Again DY = AY in the rt. angled A ADC (Ex. 2];
.. LYDA= L YAD. Similarly LZDA= ZAD; : _ZDY = L BAC. 5. Let Ad be the perp. on the hypotenuse. Then
- DAC = compt. of _ DCA = L ABC. 8. This follows at once from Ex. 7 (ii) and Ex. 3, p. 59.
9. The _ at B = diff. of _ ' BCD, BAC (1. 32], and
- at F = diff. of 3 $ FCD, FAC
= half diff. of _S BCD, BAC (hyp.). 10. Let LB be double of LA. Let cd be drawn from rt. L to middle pt. of hypotenuse AB. Then since CD = DA, LCDB is double of į CAD. .. LCBD = L CDB, so that CB= CD = half hypotenuse (Ex. 2, p. 100].
11. Let ABCD be a parm. and let BE = DF on the diag. BD. Then AS ABE, CDF are identically equal (1. 4), so that AE = FC. Also L CFE = suppe. of _ DFC = suppt. of į AEB= L AEF. .:. FC is par. to AE; :. AF is equal and par!. to EC (1. 33). 12. The As ACZ, ABX are identically equal (1. 4].
.. LZAR = LACZ. Now ext. . PRQ = sum of <S RAC, ACR = sum of LS RAC, ZAR = 2 BAC. Similarly each of 48 of APQR may be proved equal to the angle of an equilat. A. 13. The AS APS, CRQ are identically equal [1. 4].
.. PS = QR, and LAPS = QRC. Again
L APR = alt. - CRP; .. SPRE
= L PRQ. That is, SP is equal and par?. to QR. Hence SR is equal and par. to PQ (1. 33).
14. It may be proved as in the prop. that A ABF is equilat. .. L CAF = two-thirds of 2 rt. 49, and LPAF is one-third of 2 rt. <s. Again AP = AF. .. LAPF = AFP, and each is one-third of 2 rt. 48 [1. 32]. Similarly BFQ is one-third of 2 rt. 28. Thus the three _ s at F together = 2 rt. 28. Also _ s at P and Q being each equal to LC, ACPQ is equilat.
15. Let AB be the given st. line, P and Q the given points. At A and B make 48 BAC, ABD each equal to the angle of an equilat. A. Through P and a draw st. lines par! to AC, BD meeting AB in X and Y and intersecting in Z. Then XYZ is the required A.
Let o be the given pt., AB and CD the two given st. lines of which AB is the nearer to O. Draw OEF perp. to AB, CD respectively, and og perp. to OF making OG equal to OF. Draw GH perp. to AB; join Oh, and draw OK perp. to OH meeting BC in K. Then As OHG, OFK are identically equal [1. 26], and OH=OK.
The line og may be drawn par!. to AB in either direction; thus there will be two solutions corresponding to each position of o.
17. Let AB, AC, AD be the three given st. lines. Take any pt. P in AD; draw PQ par! to AB meeting AC in Q, and draw QR par. to AD. Then APQR is a par". and its diagonals bisect each other [Ex. 5, p. 64]. Thus PR is bisected by AQ. As P
be taken anywhere on Ad the number of solutions is unlimited.
18. Let L, M, N be the three given lengths, and B the given point. From B draw BC equal to N; and on BC describe a ABFC, having BF equal to twice M and CF equal to L. Bisect BF at E. Join CE, and produce it to A, making EA equal to CE. Join BA. Then BA, BE, BC are the required lines. For BC=N, and Be= M by constr., and it may be shewn that the AS AEB, CEF are identically equal (1. 4]. :. AB = CF = L. Also AE= EC by constr.
19. Let ABC be an equilat. 4; bisect the angles at B and c by BO, CO; through o draw OD, OE par!. to AB and AC respectively meeting BC in D and E. Then by 1. 29, 32 AODE is equiangular to A ABC, so that ODE is equilat. Again
- DOB = alt. L ABO= LOBD;
.. OD = BD. Similarly
OE = OC.
20. Bisect ABC by BO meeting AC in 0; through o draw OD par. to AB and OE par. to BC meeting BC and AB in D and E respectively. Then as in Ex. 19, OD= BD, and OE = BE. Hence it easily follows that EBDO is a rhombus (1. 34].
ON THE CONSTRUCTION OF TRIANGLES WITH GIVEN PARTS.
2. Let the given diff. be equal to AB and the hypotenuso equal to k. From A draw AE making with BA produced an 2 equal to half a rt. L. From centre B, with radius equal to K, describe a circle cutting AE or AE produced in the points C, c'. From C and c' draw perps. CD, c'D' to AB; and join CB, C'B. Then either of the As CDB, C'D'B will satisfy the given conditions. (See Note to Ex. 1.]
3. See fig. on p. 89. Let AB be the given sum, then using the construction and proof given on p. 90, it is shewn that AX = XC, and BX = BC. Thus CBX is the required triangle.
5. This is clearly a particular case of the preceding example.
6. Let P, Q be the given pts. through which the sides are to pass, XY the st. line in which the base is to be. At X draw XZ equal to the given altitude. Through z draw ZKL par!. to XY. Then by Ex. 7, page 49, draw through PQ two lines. PK, QK making equal angles with ZKL. Produce KP, KQ to meet XY in M and N; then KMN is the required A.
7. Draw AB, CD two parł
. st. lines at a dist. from each other equal to the given altitude. At P, any pt. in AB, make LAPQ=one-third of two rt. %$ [1. 1] and BPR = one-third of two rt. $ If PQ, PR meet CD in @ and R respectively, PQR is the required A.
8. Let AB be the base and k the given difference. Bisect AB at X; from X draw XH perp. to AB, making XH equal to k; join AH. At the pt. A make Z HAC equal to LAHX, and since HX <AX [Ex. 7, p. 38], :: C must fall on the side of AB which is remote from H. Let AC meet HX produced in C; join CB. Then ACB is the required A. (See proof of Ex. 1, p. 88.]
9. Let AB be the base; make .BAX equal to given , and AX equal to sum of sides. Join BX. X Mon centre X with med Xe describe a girale qutting AK te . Than ACB is the required A.
X M B make the - X BC ACXB. Pel BC mul
AX in C.
10. Let AB be the base; make BAX equal to given L, and AX equal to diff. of sides. Join BX. Produce AX to C, and make _XBC equal to BXC. Then ACB is the required A.
12. Let AB be the given base, k the sum of the remaining sides and x the difference of the Ls at the base. Make the LABD equal to half the LX; draw BE perp. to BD, and from centre A and with radius equal to k describe a circle cutting BE in E. At B make ZEBC equal to LAEB. Then ACB is the required A. [Ex. 7, p. 101.] Since, if AE meets BD at F, it may be shewn that CB=CF.
13. Let Ad be the given perp. and let the two given differences be X and Y. On AD as base describe a A ABD having LADB a rt. and the diff. of AB and BD equal to X. the other side of AD describe a AADC having LADC a rt. and the diff. of AC and DC equal to Y. [Ex. 10, p. 108.]
Then ABC is the required A.
VIII. ON AREAS. Page 109.
1. Let ABCD be a parm., o the middle pt. of the diag. BD. Draw any line through o meeting AB, CD in E and F respectively. Then A EOB, DOF are identically equal (1. 29, 26).
.. AEFD = A ADB = half the parm.
2. Join the given pt. to the middle pt. of a diagonal, and produce it to meet two of the parallel sides.
Examples 3 and 4 are particular cases of Ex. 1.
5. Let EXF drawn par. to AD meet DC in E and AB in F. Then AS BXF, XEC are identically equal [1. 29, 26). .. the area of the trapezium is equal to that of parm. ADEF.
6. In the preceding Example DE = half the sum of de and AF, that is half the sum of DC and AB, since BF = EC.
7. For A AXD is half the par". ADEF in Ex. 5.
8. Let E, F be middle pts. of AB and DC; join ED, EC. Then A $ EDF, EFC are equal, and A8 AED, BEC are equal (1. 38).
9. The As ADB, BCD are equal (1. 37). Take away the common part, the ABXD.