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6. Here the OCB > the 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 ▲ 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.

11. Let the bisector of A meet BC in X; then AXC is greater than XAB, that is, than ▲ XAC; .'. AC > CX.

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Similarly

BAD,

BDX is greater than
CAD. .. BDC is greater than

13. Let O be the pt.; then by 1. 21,

OA + OB < CA + CB,

OB + OC < AB+ AC,

OCOA BC + BA;

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

See fig. to Prop. 22. centres F and G draw lines; let these meet at K.

Page 40.

Let FG be the given base. Then with s with radii equal to the two given st. 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 FGFK + KG the two circles would lie wholly outside each other.

Page 44.

Here BX = XC and XA is common to the two ▲ AXB, AXC; AXB is greater or less than 4 AXC according as AB> or <AC [1. 25], and the required result follows by 1. 13.

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1. By hypoth.

Page 49.

XBC=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 = < YBC, and BC is common; .. ▲ 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 ▲ AOP, AOQ are clearly equal in all respects [1. 26];

... OP=OQ.

4. Here angles at O are equal [1. 15]; AXO = BYO, being ; and AO OB; .. ▲ AOX, BOY are equal in all respects

rt.
[1. 26].

5.

=

Follows at once from 1. 26, since in the two ▲s 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 Also let PD be nearer to PC than PE. Then PD PC

and E.

[I. 19].

>

In

Again PEC is acute, and PDE is obtuse; .. PD <PE. the same way it 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 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 I. 26.

If AB is par1. 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 draw perps. AP, BQ to it. Then AAPC, 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 ▲ AOC, BOD are equal in all respects [1. 4].

and these are alternate.

.. LOAC = L OBD,

4. Let PQ, QR be par1. to AB, BC respectively. Join BQ and produce it to O. Then PQO = int. opp. ABQ, and

opp. CBQ.

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RQO

=

int.

Hence the sum or diff. of PQO, RQO = the sum or diff. of LS ABQ, CBQ; .. PQR ABC. Similarly for the other angles.

Page 57.

1.

Let PQ drawn par1. to base BC cut the sides in X and Y. Then 4 AXP, AYQ are respectively equal to the alt. ACB, and these are equal since the A is isosceles.

OB.

ABC,

2. Let AOB be bisected by OP, and from P draw PQ par1. to Then QPO alt. ▲ POB = L POQ.

=

3. Let O be the given pt., AB the given st. line; at B make ABX equal to given . From O draw a line par1. to BX.

4. Let AD be drawn perp. to BC.

[1. 26], and is also par1. to XYZ.

Then AD bisects BAC

.. LZYA = L BYX = L Bad = dac = LYZA.

5. Let BA be produced to D, and let AX bisect DAC, and be par1. to BC. Then ext. DAX

int. opp. 4 ABC, and

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1. (i) Let AD be par1. to base BC; then ▲ BCA = CAD... the three s of A ABC

[1. 29].

=

< " CBA, BAD, which are equal to 2 rt. L

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and

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

AXC = sum of int. S XBA, BAX;

ext.

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Thus the three of ▲ 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 = <

LYCA =

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the three

3. Let XP be

XP, AQ meet at O.

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of together with BAC.

perp. to AP, and XQ perp. to AQ; also let Then AOP XOQ [1. 15];

=

rt. APO rt. 4 XQO;

.. ▲ PAO = ▲ QXO. [1. 32.]

4. Let AABC be rt. angled at C, and let D be the middle pt. of the hypot. AB.

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Also from the ▲ EDC, FDC, it may be shewn that DF

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Hence the ▲ DEA, DEC are identically equal [1. 4].

6.

.. _ DAC = 2 DCA, and hence ▲ DCB = ▲ DBC.

= EC

Let ABC be the rt. L. On BC describe an equilat. A BDC; bisect DBC by BE; then ABC is trisected by BD, BE. For DBC= two-thirds of a rt. ≤ [1. 32], .'. ▲ DBE, EBC are each one-third of a rt. .

7.

Let the bisectors be BO, CO; then BOC = suppt. of sum of OBC, OCB = suppt. of ▲ ABC.

8

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8. Let ABCD be the quadril. with at D and C bisected by DO, CO.

Then twice the sum of

Doc, odc, OCD=4 rt. "

the sum of the four angles of the quadril. Hence 2 DOC = the sum of Sat A and B.

Page 61.

S

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1. (i) The six are together equal to 8 rt. 4 [1.32, Cor. 1].

.. each = rt. 2.

(ii) The eight 3 are together equal to 12 rt. ≤ §.

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.. the exterior ▲ = 2 rt. L S rt. ≤ = rt. L.

3. Take the fig. on p. 60.

Join DA, DB; then the five-sided Thus the interior angles are together

L

S

8

10 rt.

fig. is divided into 3 ▲3. equal to 6 rt. ¿ 3. .. the int. together with 4 rt. L = Similarly the corollary may be proved for a fig. of any number of sides.

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page 60. Let AB, DC meet in G; BC, DE, BA in L; EA, CB in M. Then by

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5. Take the fig. of ED in H; CD, AE in K; Prop. 32, Cor. 2 the base to 8 rt. S. And the sum of all the of these ▲s 8 S as many rt. the fig. has sides. .. the together with 8 rt. $ = twice as many rt. sides.

of the exterior ▲ are together equal

S twice as

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s at the vertices

L

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The sum of each pair of adjacent is equal to 2 rţ. ≤3.

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CD, and BC = AD, the ▲ BCD, BAD Cor.]. .'. _ ABD = ▲ BDC; .'. &c. and ▲ ABC = ▲ ADC. .. sum of 2 adjacent = sum of of fig. = 2 rt. ▲3; .. the opposite sides are par1.

3.

▲ DAB

=

Also by Ex. 1 it is rect

4. By Ex. 2 the fig. is a parTM. angular, and since it is equilat., it is a square.

5. In fig. on p. 63 let AD meet BC in O. Then

▲ ABO = alt. ▲ OCD, LAOB = ▲ COD, and AB = CD.

.. ▲3 AOB, COD are equal in all respects.

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