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6. Here the LOCB > the L OBC; .. OB > OC.

7. BC is less than BA and AC together; take AC from both, then diff. of BC and AC is less than BA.

8. Let ABCD be a quadril. whose sides BA, CD meet in O. Then AO and OD are together > AD. .. the sum of OA, AB, BC, CD, OD > the sum of AD, AB, BC, CD.

That is, perim. of A OBC > perim. of quadril.

9. Let o be the pt.; then AO + OB > AB; OB + OC > BC; OC+OA > CA.

Hence twice the sum of OA, OB, OC > the sum of AB, BC, CA. 10, Let ABCD be the quadril.; BD, AC its diags.; then

AB + BC > AC; AD + DC > AC. Hence perim. > twice diag. AC. Similarly perim. > twice diag. BD. That is, twice perim. > twice sum of diagonals.

Let the bisector of A meet BC in X; then . AXC is greater than XAB, that is, than _ XAC; .. AC > CX. Similarly AB > BX. i. AB + AC> BX + XC.

12. Produce AD to X. Then BDX is greater than _ BAD, and – CDX is greater than CAD. .. BDC is greater than BAC. 13. Let o be the pt.; then by 1. 21,

OA + OB < CA + CB,
OB + OC < AB + AC,

OC + OA < BC + BA; .. by addition, twice the sum of OA, OB, OC <twice the sum of AB, BC, CA.


Page 40. See fig. to Prop. 22. Let FG be the given base. Then with centres F and G draw OS with radii equal to the two given st.

let these meet at K. Then KFG is the required A. If FK > FG + GK, then FK > FH, and the circle with centre F would fall outside the other circle, and there would be no pt. of intersection K. Similarly if GK > GF+ FK. If FG > FK + KG the two circles would lie wholly outside each other.

lines ;

Page 44 Here BX = XC and XA is common to the two AS AXB, AXC; :: LAXB is greater or less than 2AXC according as AB> or <AC [1. 25), and the required result follows by 1. 13.

Page 49.

1. By hypoth. LXBC= _ YCB, ?XCB= _ YBC, and BC is common; .; AXBC = A YBC in all respects [1. 26].

2. Let BX, CY be perps. to AC, AB. Then / BXC= BYC, _XCB = LYBC, and BC is common; .. A$ BXC, BYC are equal in all respects (1. 26].

3. Let o be any pt. on bisector of BAC; OP, OQ perps. on AC, AB; then As AOP, AOQ are clearly equal in all respects [1. 26]; .. OP=OQ.

4. Here angles at O are equal [1. 15); LAXO = BYO, being rt. 28; and AO = OB; .. As AOX, BOY are equal in all respects [1. 26].

5. Follows at once from 1. 26, since in the two As we have two angles and adjacent side equal.

6. Let P be the given pt., AB the given st. line. Draw PC perp. to AB, and PD, PE on the same side of PC to meet AB in D and E. Also let PD be nearer to PC than Pe. Then PD > PC [1. 19].

Again 2 PEC is acute, and _ PDE is obtuse; ..PD <PE. In the same



may be shewn that PD is less than any line which is more remote from PC. If PF be drawn on the other side of PC making < CPF equal to CPD, the As CPD, CPF are equal in all respects [1. 26]. Thus PF = PD. And as before it can be shewn that PF is greater than any line nearer to PC, and less than any

line more remote.
8. Let the two intersecting lines meet at o forming

Z POQ. Bisect – POQ by ox meeting the other given st. line AB in X. From X draw XP, XQ perp. to the given lines. Then XP=XQ by

1. 26.

If AB is part to the bisector of the POQ, the pt. X cannot be found.

9. Let o be the given pt. through which the line is to be drawn, A, B the other given pts. Join AB and bisect it in c. Join oc, and from A and B draw perps. AP, BQ to it. Then A$APC, BCQ are equal in all respects [1. 26]. The solution is impossible when o is in the same st. line as AB.

Page 54. 1. The As AOC, BOD are equal in all respects [1. 4].

.. OAC = L OBD, and these are alternate.

4. Let PQ, QR be parto AB, BC respectively. Join BQ and produce it to 0. Then < PQO = int. opp. Z ABQ, and L RQO = int. opp. L CBQ.

Hence the sum or diff. of _$ PQO, RQO = the sum or diff. of * AB, CBQ;.. PQR - ABC. Similarly for the other angles.

Page 57. 1. Let PQ drawn part. to base BC cut the sides in X and Y. Then <* AXP, AYQ are respectively equal to the alt. _ $ ABC, ACB, and these are equal since the A is isosceles.

2. Let 2 AOB be bisected by OP, and from P draw PQ par!. to OB. Then LQPO = alt. - POB= L POQ.

3. Let o be the given pt., AB the given st. line; at B make · ABX equal to given L. From o draw a line par. to BX.

4. Let AD be drawn perp. to BC. Then AD bisects L BAC [1. 26], and is also par. to XYZ.

..L ZYA=

= L BYX = L BAD= L DAC= LYZA. 5. Let BA be produced to D, and let Ax bisect L DAC, and be par!. to BC. Then ext. 2 DAX = int. opp. ABC, and

LXAC = alt. LACB. .. LABC= L ACB.

Page 59. 1. (i) Let AD be parł to base BC; then _ BCA = L CAD. :. the three 8 of A ABC = _ $ CBA, BAD, which are equal to 2 rt. 48 [1. 29].

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(ii) Let AX be drawn to a pt. in base BC; then

ext. L AXC=sum of int. _ * X.BA, BAX ; and

L AXB = sum of _ * XAC, ACX. Thus the three _ 8 of A are together equal to sum of <$ AXB, AXC, that is to 2 rt. _ $. 2. Let ABC be the A ; produce BC to X and Y ; then

LXBA=. _ S BAC and BCA;

LYCA % % CBA, BAC. :. the 2 ext. - the three < s of together with BAC.

3. Let XP be perp. to AP, and XQ perp. to AQ; also let XP, AQ meet at o. Then LAOP= _XOQ (1. 15];

rt. L APO =rt. XQO;

:: < PAO = LQXO. [1. 32.] 4. Let A ABC be rt. angled at C, and let D be the middle pt. of the hypot. AB.

Draw DE, DF perp. to AC, BC, and therefore part. to BC and AC respectively. Then in A8 AED, DFB, LADE = int. DBF.

rt. AED = rt. DFB, and AD= DB;

.. DF = AE. [1. 26.] Also from the As EDC, FDC, it may be shewn that DF = EC [1. 26].

.. AE EC. Hence the A' DEA, DEC are identically equal (1. 4].

.. DAC=L DCA, and hence DCB= DBC. 6. Let ABC be the rt. On BC describe an equilat. ABDC; bisect 2 DBC by BE; then · ABC is trisected by BD, BE. For 4 DBC=two-thirds of a rt. — [1. 32], :: 48 DBE, EBC are each one-third of a rt. L. 7. Let the bisectors be BO, CO; then - BOC =

suppt. of sum of _ S OBC, OCB = suppt. of LABC.

8. Let ABCD be the quadril. with _ 8 at D and C bisected by DO, CO. Then twice the sum of _ S DOC, ODC, OCD=4 rt. L

$ -- the of the four angles of the quadril. Hence 2 L DOC = the sum of <s at A and B.

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Page 61. 1. (i) The six _ s are together equal to 8 rt. 28 (1. 32, Cor. 1].

.:. each <=

rt. L.
(ii) The eight < s are together equal to 12 rt. 29.

.. each _ = irt. L. 2. The interior L = rt. L;

.:. the exterior L= 2 rt. 28 - art. 2 = f rt. L. 3. Take the fig. on p. 60. Join DA, DB; then the five-sided fig. is divided into 3 As. Thus the interior angles are together equal to 6 rt. 28 .. the int. s together with 4 rt. į 8 = 10 rt. L's. Similarly the corollary may be proved for a fig. of any number of sides.

4. The n angles + 4 rt. 8 = 2n rt. ¿ 8.
.. the n angles = (2n – 4) rt. 25,

2n - 4
.. each L= right angles.


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5. Take the fig. of page 60. Let AB, DC meet in G; BC, ED in H; CD, AE in K; DE, BA in L; EA, CB in M. Then by Prop. 32, Cor. 2 the base _ 8 of the exterior As are together equal to 8 rt. _ $. And the sum of all the _s of these As = twice as many rt. as the fig. has sides.

.:. the

at the vertices together with 8 rt. _ $ = twice as many rt. - as the fig. has sides.

Page 64. The sum of each pair of adjacent _ is equal to 2 rt. 28. .. &c.

2. In fig. ABCD, if AB CD, and BC AD, the AS BCD, BAD are equal in all respects [1. 8, Cor.]. .. LABD = BDC; .. &c.

DAB = _ BCD, and L ABC L ADC. .. sum of 2 adjacent _ 8 = 1 sum of _ s of fig. = 2 rt. 28; -. the opposite sides are par?

4. By Ex. 2 the fig. is a parm. Also by Ex. 1 it is rectangular, and since it is equilat., it is a square. 5. In fig. on p. 63 let AD meet BC in 0. Then Ź ABO

alt. L OCD, L AOB = COD, and AB=CD. :: AS AOB, COD are equal in all respects.


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