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10. The As ADB, BCD are equal, and therefore AC is par!. to BD (1. 39).

11. This may be proved by considering ABCX, DCX on a common base cx and of equal altitudes [Ex. 13, p. 64). Or if the diagonals meet in o, the 18 OBX, ODX are equal, and AS ODC, OBC are equal (1. 38). 12. ARBC = A QBC (Ex. 2, p. 96 and 1. 37];

.. ARBX AQCX. Again A ABQ= ABQC; .. by taking away the equal AS RBX, QCX, the area AQXR = ABXC.

13. Let ABCD be the quad'., and P, Q, R, S the middle points of the sides AB, BC, CD, DA.

Draw AC, BD intersecting at o. Let Ao meet PS at X. Then PS is par. to BD, and AX = XO. [Ex. 1 and 2, p. 96.]

First shew that the perps. from A and o on Ps are equal [1. 26). Hence it follows that the AAPS = the APOS. Similarly ABPQ= A POQ, AQCR= A QOR, and A SDR = A SOR. Hence by addition the parm. PQRS is one half of the quad'.

14. Let C and D be vertices of two equal As ACB, ADB on opposite sides of the common base AB; let CD meet AB or AB produced in E. Then if DF, CG are drawn perp. to AB, DF = CG, and the AS DEF, CEG are identically equal (1. 26].

15. Let ABCD be the trapezium having AB par. to DC. Bisect BD, CA in E and F and join EF. Then 48 AEB, AFB are equal, being halves of equal A ADB, ACB (1. 38).

:. EF is par'. to AB [1. 39).

16. (i) Let ABC, DBC be two os on a common base BC, ABC having the greater altitude.

Draw BX perp. to BC, and through A and D draw AA', DD' par'. to BC to meet BX at A', D'. From A'x cut off A'E equal to BD', and join EC, A'C, D'C.

Then EB is the sum of the altitudes of the As ABC, DBC.
And since EA' = D'B, . A EA'C= A D'BC (1. 38).

Hence A ABC + A DBC = A A'BC + A D'BC

= AA'BC + A EA'C

- AEBC. (ii) may be proved by a similar method.

17. If o lies between the parallel sides AB, CD, the perp. EOF to these sides is equal to the perp. from A to CD.

Thus the As OAB, OCD, ADC have equal bases, and the altitude of ADC is equal to the sum of the altitudes of the other two.

.. the sum of the A' OAB, OCD= AADC which is half the parm.

If o does not lie between AB and CD, the diff. of the As OAB, OCD AADC.

18. (ii) If o is within 4 DAB and its opp. vert. L, then AO intersects the parm. ; so that the perp. from c on OA is equal to the diff. of the perps. drawn from B and D to OA (Ex. 20, p. 99). Therefore since the As OAC, OAD, OAB are on the same base on, AOAC = diff, of ASOAD, OAB.

19. Let the lines EOF, GOK drawn through o parł. to AD, AB respectively meet AB in F, AD in G.

Then parm. GB= 2 A AOB, and parm. DF = 2 A DOA. And since parm. GF is common to these two parms., the diff. between parms. BO and DO = twice the diff. between As AOB and DOA = 2 AAOC (Ex. 18. ii.].

20. Draw BO part. to the diag. AC, and co parl. to AB; then ABOC is a parm. Also the perp. from Don Bo is equal to the sum of perp. from D on AC and perp. from B on AC.

.. ADBO = ADAC + ABC, since these As have equal bases (Ex. 16 (1)]. 22. Let ABC be the given A, and B the given L.

In BA take BD equal to the base of the required 4, and by Ex. 21 draw through D a st. line to meet BC produced in E, so that ADBE may be equal to A ABC.

23. Let CAB be the given A on base AB. Through a draw AD perp. to AB and equal to the given altitude, and through C draw CE parl

. to AB meeting AD in E. Join DB, and draw EF par!. to DB meeting AB in F. Then A DAF = A EAB = A CAB [Ex. 21]. 24. Let AB be the given base, CD the given line in which the vertex is to lie. On AB describe a AABE equal to the given A[Ex. 21]. Through E draw EC par! to AB meeting cd in C; then CAB is the required A.

25. (i) On AB the given base describe A ABC equal to the given [Ex. 21, p. 111]. Bisect AB at D, draw DE perp. to AB meeting CE, drawn parl. to AB in E; then AEB is the reqa. A.

(ii) Draw AF perp. to AB meeting CF, drawn par!. to AB in F; then FAB is the req". A.

26. Let X, Y be the two given A$. If they are not on equal bases, make a triangle z equal to Y and having a base equal to that of x [Ex. 21, p. 111]. The construction now follows easily by Ex. 16 (i), p. 110.

27. Through A draw AD part. to BC meeting BX in D. Then ACDB= A ABC. Through x draw XF to meet BC so that

AXBF = A CDB (Ex. 21, p. 111]. 28. The As BDX, BDC are equal (1. 37]. To each add A ABD.

29. Take a five-sided figure ABCDE. Join EC. Through D draw DF par!. to EC meeting BC at F. Then the quadrilat. ABFE = the given fig. For AEFC = A EDC (1. 37]; add to each the fig. ABCE. Similarly a six-sided figure can be replaced by an equal figure of five sides, and so on. Thus any rectilineal figure can ultimately be reduced to a triangle of equal area.

30. Through C and D draw CE, DF par to BX and AX respectively meeting AB in E and F; then EXF is the reqd. A [Ex. 21, p. 111].

31. Let ABCD be the parm.; through c draw CE parl. to the diag. BD; bisect BD at F, and draw FG perp. to BD meeting CE in G; join GB and GD. Then it is easily seen that BGD is an isosceles A equal in area to ABCD, and by drawing a A BHD identically equal to ABGD but on the opp. side of BD, a rhomhus BGDH will be formed equal to the given parm.

32. Let ABC be the given A on BC as base. Bisect BC in D, and draw AE par?. to BC. With centre B and radius equal to half the sum of sides BA, AC, describe a circle cutting AE in E;

through D draw DF par!. to be meeting AE in F. Then par”. eEFD is double of ABD, and is therefore equal to A ABC. Also sum of BE and DF =sum of BA and AC, and sum of BD and EF = BC.

33. The bisecting line is the median through the given angular pt. (1. 38).

34. Join the given pt. to the pts. of trisection of the opp. side.

35. Divide the side opp. to the given pt. into the required number of parts, and join the points of division to the given pt.

38. As the method is quite general it will be sufficient to take a particular case. Let ABC be a triangle from which it is required to cut off a fifth part by a st. line through a pt. X in AB. Take BD a fifth part of BC [Ex. 19, p. 99). Join AD, and through X draw Xe to meet BC in E, so that ABXE= A ABD [Ex. 21, p. 111].

40. With the construction of Ex. 28, p. 112 let BAX be a A equal to the given quadrilateral. Take Ay equal to one-fifth of AX; join BY. Then A BAY = one-fifth of A BAX, that is, of the quadrilateral. The method is quite general.

sq. on AE

41. (i) Let Al meet BC in X.

sq. on AB=sum of sqq. on BX, AX

=sum of sqq. on DL, AX;
=sum of sqq. on AL, LE

=sum of sqq. on AL, CX. i. sum of sqq. on AB, AE = sum of sqq. on DL, AL, AX, CX = sum of sqq. on AD, AC.

(ii) Produce AC to M making CM equal to AC; join BM. Then _ BCM = suppt. of - ACB = Z ECK, and A8 BCM, ECK are identically equal (1. 4). Therefore sq. on EK = sq. on BM = sum of

sqq. on BA, AM (1. 47]
= sq. on BA with four times sq. on AC.

(iii) Sq. on EK = sq. on AB, with four times sq. on AC,

sq. on FD = sq. on AC, with four times sq. on AB; .. sum of sqq. on EK, FD=five times sq. on AB with five times sq. on AC

= five times sq. on BC. 42. Let AB be divided at C; from c draw CD perp. to AB and equal to CB. Join AD, DB. Then LCDB = LCBD (1. 5]; i. - ADB is greater than 2 ABD, and .. AD<AB. Now sq. on AD = sum of sqq. on AC, CD [1. 47]

= sum of sqq. on AC, CB.
.. sq. on AB is greater than sum of sqq. on AC, CB.

43. Let sq. on AC be greater than the sum of sqq. on AB, BC; then shall ABC be obtuse. Draw BD perp. to BC and equal to AB.

Then sq. on DC = sum of sqq. on DB, BC (1. 47].
But sq. on AC > sum of sqq. on AB, BC.

.. sq. on AC > sq. on DC, or AC > DC. Hence by i. 24 in the As ABC, DBC, the LABC is greater than 2 DBC; that is, LABC is obtuse.

44. The sq. on BQ =sum of sqq. on AB, AQ [1. 47]; and

sq. on CP = sum of sqq. on AC, AP. i. sum of sqq. on BQ, CP: = sum of sqq. on AP, AQ, together with sum of sqq. on AB, AC.

That is, sum of sqq. on BQ, CP = sum of sqq. on PQ, BC.

45. Let the medians be BQ, CP. Then by preceding example four times the sum of the sqq. on BQ, CP= four times the sum of the sqq. on PQ, BC.

But four times the sq. on PQ=sq. on BC (Ex. 3, p. 97]. .. four times the sum of the sqq. on BQ, CP = five times the sq. on BC.

46. Let AB be a side of the given square; produce AB to C making BC equal to AB. On AC describe an equil. A ACD.

Join DB. Then, as in 1. 11, DB is perp. to AC, and therefore the sq. on AD= sq. on AB with sq. on DB. But AD is double of AB; i. sq. on AD=4 times sq. on AB.

sq. on DB = 3 times sq. on AB.

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