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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,
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.
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.
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.
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
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 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].
(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.
Page 61. 1. (i) The six _ s are together equal to 8 rt. 28 (1. 32, Cor. 1].
.:. each <=
.. 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.
2n - 4
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.
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.