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4. Let BO, CO drawn from extremities of the base meet in O. Then since OB OC, .. OBC= OCB. And ▲ ABO ▲ ACO. ABC = LACB.

..

8

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5. Let BD, CE be the equal perps. drawn from ends of base Then in ▲ ABD, ACE, BD = CE, LADB = AEC [Ax. 11] and the at A is common; .. AB AC [1. 26].

BC.

6. Let CD meet AB in E. < BCD = < BDC [1. 5]. .. LACB equal [1. 4].

.. CAB = L DAB.

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Now in ▲ CAE, DAE, CAE = L DAE, LACE = LADE, and AE is common. .. LAEC = LAED [1. 26].

7. In the fig. on p. 15 let BY, CX be perp. to sides and intersect in O.

The As CBX, BCY are equal in all respects [1. 26];

.. XCB= _YBC; .. OB = OC [1. 6]; ..A AOB, AOC are equal in all respects [1. 8].

S

8. In ▲ BAD, DAE, BAD = L DAE, ▲ BDA = LADE, and AD is common. .. BD = DE [1. 26].

9. By hypothesis AB AD, BC=CD, AC is common; .. ▲ ABC, ADC are identically equal [1. 8].

10. The ▲ ABD, BAC are equal in all respects [1. 8].

.. LABD = L BAC.

.. ▲ AKB is isosceles [1. 6]. Similarly ▲ KDC is isosceles.

11. Here the greatest angle is a right angle [1. 32]. Hence the required result follows by Ex. 4, on p. 59.

II. ON INEQUALITIES. Page 93.

1. See Solution of Ex. 5, p. 38.

2. See Solution of Ex. 11, p. 38.

3. See Solution of Ex. 6, p. 49.
4. See Solution of Ex. 8, p. 38.
5. See Solution of Ex. 13, p. 38.

6. See Solution of Ex. 10, p. 38.

7. Let O be the given pt., AC and BD the diagonals; then AO + OC > AC, BO+OD > BD [1. 20]. The exceptional case is when O is at the intersection of the diagonals.

8. Let the median AD bisect BC; produce AD to E making DE equal to AD, join EC. Then ▲ ABD, EDC are identically equal [1. 4] and AB = CE. Now AC+ CE > AE [1. 20].

That is, AB+ AC > 2AD.

9. This follows at once from Ex. 8, since twice the sum of the sides is greater than twice the sum of the medians.

10. Let the median AD bisect BC. If AD > DC, LACD is greater than DAC; similarly DBA is greater than DAB. Hence the sum of the angles at B and C is greater than the angle at A; that is, BAC is acute [1. 32]. The other cases follow

similarly.

11. In the rhombus ABCD let ▲ DAB be greater than ABC. Then since the sides of a rhombus are equal it follows that

DB > AC [1. 25].

13. In the fig. on p. 94 let AD be perp. to BC.

Then

and

but

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▲ DAC = compt. of

ACD,
ABD;

DAB = compt. of

ACD is greater than ABC;

DAC is less than DAB.

.. BAD is greater than half vert. BAC.

.. AD lies within the PAC.

Thus by Ex. 12, AP lies between AD and AX, and by Ex. 3 it is intermediate between them in magnitude.

III. ON PARALLELS. Page 95.

2. From O any pt. on the bisector of

BAC draw OP par1. to

AB, and OQ par1. to AC. Then QOA = LOAP = LOAQ.

.. QO = AQ=OP since OPAQ is a parm. Also OQ = AP; thus the fig. is equilat.

3. Let D be the pt. of intersection of AB and CD; then XYD = alt. LYDAYDX.

.'. YX = DX = XZ similarly.

4. See Ex. 4, page 54.

5. Let POQ be terminated by the parls. at P, Q, and bisected at O; through O draw XOY perp. to the par1.; then ▲3 XOP, YOQ are identically equal [1. 26].

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6. The two formed are identically equal [1. 29, 1. 26].

7. Let O be pt. equidist. from the par1s., and let POQ, XOY be drawn to cut them. Draw LOM perp. to the pars; then ▲ LOP, MOQ are identically equal [1. 26]. .. OP=OQ; similarly OX = OY; hence PX = QY [1. 4].

8. Draw XP perp. to CD; bisect BXP by XQ meeting CD in Q. Through a draw QY par1. to XP meeting AB in Y; then Y is the required pt. For

< YXQ = < QXP = alt. ▲ XQY;

.. QY = XY [1. 6].

9. Bisect ACB by CD meeting AB in D; draw DE par1. to

BC meeting AC in E.

Again ext. 4 ADE

[1. 29]; .. AD AE.

BC.

Then EDC alt. DCB=

=

=

.. EC = ED [1. 6].

= 4 DCE.

int. opp. ABC ACB = ext. 4 AED Hence BD = EC = ED.

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draw DOE par1, to

10. Bisect the ABC, ACB by BO, CO;
Then as in preceding examples it easily follows that

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11. Produce BC to F. Bisects ACF, ABF by CO, BO. Draw OED par1. to BC meeting AE in E and AB in D. Then as That is, DE is the diff. between BD

before DO

and CE.

=

BD, EO EC.

=

IV. ON PARALLELOGRAMS. Page 97.

3. In Ex. 2 it is shewn that BC = ZV, and that YZ = YV. Thus BC= 2ZY.

H. K. E.

2

4. In the fig. of Ex. 1 let X, Y, Z be the middle pts. of the sides. Then ZY is par1. to BX; similarly XY is par1. to BZ ; ..BZYX is a parm., and its diag. ZX bisects it.

5. In fig. of Ex. 2 let ADE be any line meeting ZY in D and BC in E. Then in ▲ ABE, ZD bisects AE [Ex. 1].

Then

6. In fig. of Ex. 1 let X, Y, Z be middle pts. of sides. Through X, Y, Z draw BC, CA, AB respectively par1. to YZ, ZX, XY. by 1. 34, AZ = XY = BZ.

7. Through P draw PQ par. to AC meeting AB in Q; on QB make QX equal to AQ; join XP and produce it to meet AC in Y. Then QP drawn from middle pt. of AX par1. to AY bisects XY [Ex. 1].

8. Let AC meet BX in E and DY in F. Then DY is par1. to XB [1. 33], therefore by Ex. 1, CE is bisected by YF, and AF is bisected by XE.

9. Let P, Q, R, S be middle pts. of sides AB, BC, CD, DA respectively. Then by Ex. 2, PQ and SR are each par1. to AC, and PS and QR are each par1. to BD.

10. In last Ex. PR and QS are diags. of a parm, and therefore bisect each other [Ex. 5, p. 64].

11. Let BC, AD be the oblique sides; join BD. Let X, Y be middle pts. of BC, BD; then XY is par!. to DC [Ex. 2]. Also XY produced bisects AD [Ex. 1]. Similarly for the other diagonal.

12. As in Ex. 11, let X, Y, Z be middle pts. of BC, BD, AD; then XY = half CD, and YZ half AB [Ex. 3]. Again, if XYZ meets AC in P, XY = half CD, and XP = half AB; .. PY = half diff. of AB and CD.

14. Let three par1. st. lines meet a fourth st. line in A, B, C making AB equal to BC, and let them meet another st. line in P, Q, R. Through P draw PST par1. to ABC meeting QB in S and RC in T. Then PS = AB = BC= ST [1. 34]; hence PQ = QR [Ex. 1]. 15. Let AB, CD be equal and par1. st. lines and let XY, PQ be their projections on any st. line; let AE, CF drawn par1. to XY meet BY, DQ in E and F respectively. Then ▲ ABE, CDF are identically equal [1. 26], so that AE = CF. .. XY = PQ [I. 34].

S

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Let OZ be perp. to XY. Then XZ ZY being projections .. the ▲ XZO, YZO are identically

16. of the equal lines AO, OB. equal [1, 4].

17. Draw ALM par1. to XY meeting OZ in L, BY in M. Then BM=2OL [Ex. 1, 3, p. 96].

Also AX = LZ

= MY.

.. 2OZ = 2OL + 2LZ

= BM+MY+AX

=

BY + AX.

18. The first case can be proved as in Ex. 17. In the second case, with same construction as before, BN = OL = NM.

.. 2oz = 2OL - 2LZ = BM - MY-AX = BY -- AX.

20. Let ABCD be the given parm. Through A draw any st. line EAF and let CX, BF and DE be perps. on this line. Through C draw CH par1. to EF meeting FB in H. Then it is easily seen that ▲ BCH, DAE are identically equal [1. 26]; .. BH =

.'. DE + BF = BH + BF = CX [1. 34],

for CXFH is a parm. by constr.

DE.

21. Let AX, CY be perps. on the given line from one pair of opp., and DP, BQ perps. from the other pair of opp. 3. Let the diagonals intersect in E, and let EF be perp. to the given line. Then AX + CY = 2EF [Ex. 17, p. 98]

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since E is the middle pt. of the diagonals [Ex. 5, p. 64].

22. From D in base BC let DE, DF be drawn perp. to AC, AB respectively; from B let BG be drawn perp. to AC. Draw BH par1. to AC to meet ED produced in H. Then GH is a parm, and BG EH. Also ▲ BFD, BHD are identically equal [1. 26], so that DH DF. That is, BG = sum of DE and DF.

8

23. Take D in CB produced, then with the same lettering and construction as in Ex. 22 it is easily seen that BG = HE difference between DE and DF.

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24. Let OX, OY, OZ be perps. to BC, CA, AB respectively. Through O draw POQ par. to BC; then APQ is an equilat. ▲ and sum of OY and OZ perp. from P on the opp. side = perp. from A on PQ since ▲ APQ is equilat. Hence sum of OX, OY, OZ = perp. from A on BC.

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