Page images

4. Let BO, co drawn from extremities of the base meet in o. Then since OB=OC, .. LOBC= = LOCB. And LABO = LACO. .. < ABC = L ACB.

5. Let BD, CE be the equal perps. drawn from ends of base BC. Then in AS ABD, ACE, BD= CE, LADB = LAEC [Ax. 11] and the L at A is common; .. AB= AC [1. 26). 6. Let CD meet AB in E. Then L ACD

ADC, and _ BCD= L BDC (1. 5). :: LACB = LADB; :: AS ACB, ADB are equal (1. 4]. iiCAB = 2 DAB.

Now in AS CAE, DAE, - CAE= 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 (I. 26];

i. _ XCB= _ YBC; .. OB = pc (1. 6]; .. AR AOB, AOC are equal in all respects (I. 8].

8. In AS BAD, DAE, 2 BAD=L DAE, LBDA= L ADE, and AD is common. .:. BD = DE [1. 26].

9. By hypothesis AB=AD, BC=CD, AC is common; .. As ABC, ADC are identically equal (1. 8]. 10. The A8 ABD, BAC are equal in all respects (1. 8].

.. ABD =

L BAC. :: AAKB is isosceles (1. 6]. Similarly A 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.


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 AS ABD, EDC are identically equal (1. 4] and AB= CE. Now AC + CE > AE (1. 2C].

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 L 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.

_ DAC = compt. of LACD, and

į DAB= comp. of LABD; but

- ACD is greater than ABC;

.. LDAC is less than [ DAB.
.. BAD is greater than half vert. – BAС.

.:. 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 sector of L BAC draw OP par. to AB, and OQ par. to AC. Then L QOAN 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

L XYD = alt. LYDA= _ YDX.

:: 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 parls.; then As XOP, YOQ are identically equal (1. 26].

6. The two A$ formed are identically equal (1. 29, 1. 26).

7. Let o be pt. equidist. from the parts, and let POQ, XOY be drawn to cut them. Draw LOM perp. to the parls; then A $ 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 par!. to XP meeting AB in Y; then Y is the required pt. For

LYXQ= L QXP = alt. - XQY;

:. QY = XY [1. 6]. 9. Bisect L ACB by CD meeting AB in D; draw DE par!. to BC meeting AC in E. Then - EDC = alt. - DCB = = L DCE.

:. EC = ED (1. 6]. Again ext. - ADE int. opp. - ABC= _ ACB =- ext. AED [1. 29]; .. AD= AE. Hence BD=EC = ED.

10. Bisect the < SABC, ACB by BO, CO; draw DOE par', to BC. Then as in preceding examples it easily follows that

DO = DB, and EO = EC. 11. Produce BC to F. Bisect 48 ACF, ABF by co, BO. Draw OED parł. to BC meeting AE in E and AB in D. Then as before DO = BD, EO = EC. That is, de is the diff. between BD and CE.

[blocks in formation]

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


4. In the fig. of Ex. 1 let X, Y, Z be the middle pts. of the sides. Then zy is par! to BX; similarly XY is par! 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 A ABE, ZD bisects AE [Ex. 1].

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

7. Through P draw PQ part. 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 par-. to AY bisects XY [Ex. 1].

8. Let AC meet BX in E and DY in F. Then DY is part. 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 par. to AC, and PS and QR are each part. 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 par!. 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 par. 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 par!. st. lines and let XY, PQ be their projections on any st. line; let AE, CF drawn par!. to XY meet BY, DQ in E and F respectively. Then AS ABE, CDF are identically equal (1. 26), so that AE = CF. :. XY = PQ (1. 34).

16. Let oz be perp. to XY. Then XZ = ZY being projections of the equal lines AO, OB. .. the A' XZO, YzO are identically equal (1. 4].


17. Draw ALM par. to XY meeting oz in L, BY in M. BM = 20L [Ex. 1, 3, p. 96). Also AX = LZ = MY.

.. 2oz 20L + 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.

.. 202 = 20L - 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 parto EF meeting FB in H. Then it is easily seen that AS BCH, DAE are identically equal [1. 26]; .. BH = DE.

:. DE + BF = BH + BF = cx (1. 34], for CXFH is a parm. by constr.

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. 45. Let the diagonals intersect in E, and let EF be perp. to the given line. Then AX + CY = 2EF [Ex. 17, p. 98]

= DB + BQ, 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 par?. to AC to meet ED produced in H. Then GH is a parm, and BG= EH. Also AS BFD, BHD are identically equal (1. 26), so that DH = DF. That is, BG = sum of de and DF.

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.

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

« PreviousContinue »