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and the contd. PQS = the contd. & PRS;

::. Δ ΡQS = Δ PRS in all respects

▲ QPS = 4 RPS.

[1. 4.]

so that

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and

.. L DAB = DCB.

9. In the A3 BCD, ADC,

BC = AD, and DC is common to both,

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and

Then each of the As ALN, LMN is isosceles,

.. LALNL ANL,

L MLN = MNL,

.. LALMLANM

[1. 5.]

KEY TO EXERCISES.

1.

BOOK I.

Page 13.

Let AB be the given line, X the line to which the sides are to be equal. From centre A with rad. X draw FCD. From centre B with rad. X draw GCE cutting the former in C. Join CA, CB. Then CAB is the A, for by constr. and def. 11 each of the sides CA, CB = X.

2. Let AB be the given line; produce AB to D making BD equal to AB, and produce BA to E making AE equal to AB. From centre A with rad. AD draw DCF; from centre B with rad. BE Then ACB is the ▲.

draw

ECG. Join AC, BC.

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[For solutions to the Exercises on Pages 17-17в see Introduction.]

Page 23.

1. The point F would fall in one of the following ways:

(1) above DE and below A, in which case the proof would still hold,

(2) on the pt. A, in which case the constr. fails,

(3) above the pt. A, in which case the constr. fails because the line AF does not fall within the given angle.

2. The ▲ DAF, EAF are equal in all respects [1. 8, Cor.].

H. K. E.

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Page 24.

1. With fig. of Prop. 10, CA = CB, CD is common, and LACDL BCD.

.. ▲ ACD, BCD are equal in all respects [1. 4].

2.

Bisect the given st. line, and with the half line for the equal sides describe an isosc. A as in Ex. 1, page 13.

Page 25.

3. Take a pt. X on CF, or CF produced. Then DC = CE, CX is common, and ▲ DCX= ECX, being rt. §. .. DX= EX.

Page 27.

1. Let A be the vertex and BC the base bisected at D. Then BD = DC, AD is common, and AB = AC. .. LBDA = CDA [1. 8].

2. Let X, Y be the middle pts. of the equal sides AC, AB. Then in ▲ BXC, CYB the sides XC, CB are equal to YB, BC respectively, and the contained ≤ are equal; .. BX = CY.

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3. Let BDCE in BC the base of isosc. A ABC. Then ▲ ABD, ACE are equal in all respects [1. 4].

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DC,

4. Let BD be a diagonal of a quadril. which has AB BC= DA. Then ▲ ABD, BCD have their sides respectively equal; .. 2 DAB = _ DCB. Similarly for the other pair of angles.

5. Here YAB = LYBA; and XAB = XBA;
.. LXAY =
= LXBY.

Join XY, then it easily follows that AXAY AXBY in all respects.

6.

ABCD is a rhombus and BD a diagonal.

CBD are equal in all respects [1. 8].

Then ▲ ABD,

7. Let BX, CY bisects ABC, ACB of isosc. ▲ ABC, and let them meet in O. Then OBC, OCB being halves of equal angles are themselves equal; .. OBC is isosceles [1. 6].

8. Here BA, AO = CA, AO respectively, and BO = OC. [Ex. 7.]

.. L BAO = L CAO.

Let D, E, F be middle pts. of BC, CA, AB respectively; then in ▲ AFE, BFD, AF = BF, AE = BD, and FAE= FBD.

.. FE = FD. Similarly DE = FE = FD.

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ABCD the rhombus has diags. BD, CA meeting at X. Then AB = BC, BX is common and LABX LCBX [Ex. 6]. .. ▲ ABX, CBX are equal in all respects [1. 4].

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12. In ▲ BAY, CAX, at A is common and BA, AY CA, AX. .. ▲3 are equal in all respects; .. LABY LACY. But LABC= = LACB; .. OBC = LOCB; that is, BOC is isosceles. Again AB = AC and AO is common, and BO = OC;

.. L BAO- = L CAO.

L BAZ =

Let AO meet BC in Z. Then BA, AZ = CA, AZ, and = ▲ CAZ; .'. ▲ BAZ, CAZ are equal in all respects. 13. AB the base, P the length of perp. Bisect AB in C. Draw CX perp. to AB and equal to P. Join AX, BX. Then ▲ ACX, BCX are equal in all respects; .. AX = BX.

14. A, B the given pts., XY the given line. Join AB, and bisect it in C. Draw CP perp. to AB meeting XY in P. Then As ACP, BCP are equal in all respects; .. AP = BP. The construction fails when CP is par1. to XY; this will be the case when AB is perp. to XY, as will be seen later.

Page 29.

1. In AABC let BC be produced both ways to X and Y. Then ABC, ACB are supplementary to equal angles, and are therefore equal.

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2. In the fig. the XOB, YOB together make up half the <S AOB, BOC; that is, half of 2 rt.

.

3. Since XOY is a rt. 4, and AOB, BOC together 2 rt. 4s;

.. AOX and COY a rt. L.

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4. The COX is supplementary to AOX, and

LAOX = L BOX.

The second case is similar.

Page 30.

By Ex. 6, p. 27, ▲ BAO = ▲ DAO.

.. ▲ BAO, DAO are equal in all respects [1. 4]; .. DOA = LAOB. Again, the ▲ AOB, COB are equal in all respects [1. 8]; .. LAOB = ▲ COB

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.. DOA, AOB are together equal to 2 rt. 8.

.. OB and OD are in one st. line.

Page 33.

= art. 4.

1. If any two st. lines would meet at a pt. A if produced, and are cut by another st. line BC, the interior angles on the same side, viz. ▲ ABC, ACB are together less than 2 rt. §.

Then

4 rt. s [1. 13]. Of

2. In the fig. to the Prop. let CB be produced to E. the DCA, ACB, CBA, ABE together these, ABC, ACB are less than 2 rt. §; .'. ≤3 ACD, ABE are

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less than ▲ AXC, AXB; that is, less than 2 rt. 3 (1. 13].

Page 38.

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1. A ▲ must have two acute 4o [1. 17]. .. the rt. is the greatest, and has the greatest side opposite to it.

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2. Let ABC be a ▲ having ABC ACB. Then AB cannot be > AC, for then the ACB would be > the ABC. Similarly AB cannot be < AC.

3. Here the ACB > the ADC [1. 16]; .. the ABC > the ADC; :'. in ^ABD, AD > AB.

4. Let ABCD be the quadril. having AB the least and CD the greatest side. Join BD. Then the ABD > the ADB because AD > AB; and the CBD > the CDB, because DC > BC. is, the whole ABC > the whole ▲ ADC.

The other case can be proved similarly by joining AC.

L

AXB > the

5. Let X be the pt. in base BC; then the AXB > the and ACX is not less than 4 ABC; .. the that is, AB > AX.

That

ACX, ABX;

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