25. Let O be the given pt., AB, CD the par1. 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 par1., 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 par1. to AB meeting PQ in Q. Then QR is the required line [1. 33].

OS.

27. Let the given lines PO, QO meet in O; bisect POQ by Draw OR perp. to OS on the same side as OQ and equal to the given length; through R draw RT par1. to OP meeting OQ in T. Through T draw TV par1. to OR meeting OP in V. TV is equal and par1. to OR [1. 33], and since it is perp. to OS it is equally inclined to OP and OQ [1. 26].

Then

28. Join AP; bisect AP in Q, and draw QR par1. to AB (the further line) meeting AC in R. Join PR and produce it to meet Then QR bisects PS [Ex. 1, p. 96].

AB in S.

V.

1.

MISCELLANEOUS THEOREMS AND EXAMPLES.

Page 100.

ACD = ADC [1. 5] and ACB: = LABC; .. ▲ BCD=sum of CBD, BDC. BCD is a rt. [1. 32].

8

That is,

2. Let CD join the rt. C to D the middle pt. of AB. Draw DE par1. to AC meeting BC in E. Then BC is bisected at E [Ex. 1, p. 96]. Also 3 DEB, DEC are equal, being rt. ≤ §.

... ▲ 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];

Similarly

ZDA

.'. 4 YDA = L YAD.

= L ZAD; .. ZDY = L BAC.

5. Let AD be the perp. on the hypotenuse.

▲ DAC = compt. of DCA = ABC.

Then

8. This follows at once from Ex. 7 (ii) and Ex. 3, p. 59.

=

10. Let B be double of A. Let CD be drawn from rt. to middle pt. of hypotenuse AB. Then since CD DA, 4 CDB is double of CAD. .. L CBD = LCDB, so that CB CD = half hypotenuse [Ex. 2, p. 100].

=

DF on the diag. BD.

FC.

11. Let ABCD be a parm. and let BE Then ▲ ABE, CDF are identically equal [1. 4], so that AE Also CFE suppt. of DFC = suppt. of is par'. to AE; .. AF is equal and par!. to EC [1. 33].

=

AEB

LAEF.

12. The AACZ, ABX are identically equal [1.4].

Now ext. PRQ = sum of LS RAC, ACR = sum of

.. FC

8 RAC,

ZAR: = BAC. Similarly each of 4 of APQR may be proved equal to the angle of an equilat. A.

S

13. The APS, CRQ are identically equal [1. 4].

Again

.. PS = QR, and APS = 4 QRC.

▲ APR alt. L CRP;

=

.. SPR: = PRQ.

That is, SP is equal and par1. to QR.

par'. to PQ [1. 33].

Hence SR is equal and

14. It may be proved as in the prop. that ▲ ABF is equilat. .. L CAF = two-thirds of 2 rt. $, and PAF is one-third of 2 rt. 4. Again AP = AF.

8

.. LAPF = AFP, and each is one-third of

2 rt. [1. 32]. Similarly BFQ is one-third of 2 rt. 4. Thus the three at F together

each equal to C, ACPQ is equilat.

15. Let AB be the given st. line, P and Q the given points. At A and B make BAC, ABD each equal to the angle of an equilat. A. Through P and Q draw st. lines par. to AC, BD meeting AB in X and Y and intersecting in Z. Then XYZ is the required A.

16.

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 A OHG, OFK are identically equal [1. 26], and

OH = OK.

The line OG may be drawn par1. 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 par1. to AB meeting AC in Q, and draw QR par1. to AD. Then APQR is a parm. and its diagonals bisect each other [Ex. 5, p. 64]. Thus PR is bisected by AQ. As P may 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 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 ▲ AEB, CEF are identically equal [1. 4]. .. ABCF: L. Also AE EC by

constr.

=

19. Let ABC be an equilat. A; 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 ODE is equiangular to AABC, so that ODE is equilat. Again

20. Bisect ABC by BO meeting AC in O; through O draw OD par!. to AB and OE par1. 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].

VII. ON THE CONSTRUCTION OF TRIANGLES WITH GIVEN PARTS.
Page 107.

2. Let the given diff. be equal to AB and the hypotenuse equal to K. From A draw AE making with BA produced an equal to half a rt. ▲. 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 par1. 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 ▲.

7. Draw AB, CD two par1. st. lines at a dist. from each other equal to the given altitude. At P, any pt. in AB, make ▲ APQ= one-third of two rt. [1. 1] and BPR = one-third of two rt. 8. If PQ, PR meet CD in Q 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 HAC equal to AHX, 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 L, and AX equal to sum of sides. Join BX. Ron centre X with md Xo dusaribe a airale qutting AK in C. Than ACB is the required A.

X MB make the L XBC = LCXB. Let BC meet

AX in C.

10. Let AB be the base; make ▲ BAX equal to given 4, and AX equal to diff. of sides. Join BX. Produce AX to C, and make LXBC equal to Then ACB is the required A.

BXC.

12. Let AB be the given base, K the sum of the remaining sides and X the difference of the S at the base. Make the

ABD equal to half the X; 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 EBC 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 CBCF.

13. Let AD be the given perp. and let the two given differences be X and Y. On AD as base describe a AABD having ▲ ADB a rt. and the diff. of AB and BD equal to X. Also on the other side of AD describe a AADC having ADC a rt. and the diff. of AC and DC equal to Y. [Ex. 10, p. 108.]

Then ABC is the required A.

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 ▲ EOB, DOF are identically equal [1. 29, 26].

... AEFD = ▲ 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.

.. the

5. Let EXF drawn par1. to AD meet DC in E and AB in F. Then ▲ BXF, XEC are identically equal [1. 29, 26]. 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 AAXD is half the parTM. ADEF in Ex. 5.

8. Let E, F be middle pts. of AB and DC; join ED, EC. Then ▲ EDF, EFC are equal, and ▲ AED, BEC are equal [1. 38]. 9. The AADB, BCD are equal [1. 37]. Take away the common part, the ▲ BXD.