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The same diagram and letters will serve for examples 6–10 inclusive.

6. AS AOB, COD are equal in all respects [1. 4]. Hence AB, CD are equal and parallel.

7. LABO = LACO being halves of equal <. .. AB= AC.

8. In AS ABD, BDC, AB = DC, BD is common, AD= BC; .. LABD= L BDC. :. each is a rt. [1. 29].

9. In AS ABD, BDC if _ S ABD, BDC are not equal, AD is not equal to BC (1. 25].

10. Let the line meet AB in P, CD in Q; then AS OPB, OQC are equal in all respects [1. 26)

... OP=OQ. 11. In parms. ABCD, EFGH, AB =

EF,

FG, and L ABC= LEFG. Then A ABC may be made to coincide with A EFG (1. 4] and A ADC will coincide with A EHG (1. 7].

13. Let AP, CQ be perp. to diag. BD. Then LADP=alt. L QBC, LAPD = L CQB, and AD = BC. .. AS APD, BQC are equal in all respects [1. 26].

14. AX is equal and part. to YC. :. &c. (1. 33].

BC =

Page 65. 1. The construction consists of three steps: (i) joining A to an extremity of BC, (ii) describing an equilat. A on the joining line, (iii) producing two sides of the equilat. A.

Now each of these steps may be performed in two ways: for (1) A might be joined to either extremity of BC, (2) the equilat. A might be described on either side of the joining line, and (3) the two sides might be produced in either direction. Hence the no. of constructions is 2 x 2 x 2, or 8.

Exceptional case, when A is situated at the vertex of an equilat. A on BC.

2. In fig. to Prop. 15 let EX, EY bisect _ S BEC, AED respectively. Then <$XEC, CEA, AEY= 4 * XEB, BED, DEY respectively. Thus sum of <$XEC, CEA, AEY = 2 rt. 28.

3. By equal 4$, BAE = alt. ECF and AB=FC. .. AF and BC are equal and par'. (1. 33]. Thus ABCF is a parm. and As ABC, AFC are equal in all respects (1. 34).

5. See solution of Ex. 4, p. 61.

6. In fig. to Prop. 21, suppose BD, CD bisect base < 5, and let Ad be produced to F; then _ BDF = sum of _ * DBA, BAD; < CDF = sum of _ S DCA, CAD. Hence LBDC = the angle at A together with half the sum of the base angles.

7. In A ABC let the external bisectors of $ B and C meet at F, and the internal bisectors at E. Then by Ex. 2, page 29 each of _ SEBF, ECF is a rt. L. 1. LF is the suppt. of LE; that is ?F= sum of _ * EBC, ECB.

8. Let the st. lines intersect at 0, and let P be the pt. from which perps. PX, PY are drawn.

If PX and PY are equal it may be shewn [1. 26] that OQ bisects both the 28 XOY, XPY.

But if px is greater than Py, let the bisector of the - XOY meet Px in Q; and let the bisector of the < YPQ meet OY at s. Draw QR perp. to OY and therefore par. to PY.

Then the AS OQR, OQX are equal in all respects [1. 26).
.. LRQO = L XQO.

But since RQ is part. to YP; .. iRQX = _ YPX; and the halves of these are equal; that is, LOQX = < SPX. .. Sp is parl. to OQ.

9. AP= AQ; .. LAPQ = LAQP; . each is half the suppt. of L BAC. Hence LAPQ= L ABC. :. PQ is par. to BC (1. 28].

10. Join AD, BE, CF; then AD is equal and par. to BE, and CF is equal and par!. to BE (1. 33). .. AD is equal and par. to CF; .. AC is equal and parl. to DF.

11. This will be easily seen from the adjoining diagram.

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12. In the quadril. ABCD, let AB be par! to CD, and AD equal to BC. Draw BE par!. to. AD; then ABED is a parm. and BE= AD = BC. .. LBCD BED= the suppt. of _ DAB. Again, since the four _ s of the fig. are together equal to four rt. 2$, the other pair of opp. _ s are supplementary. Join AC, BD; then in AS DEB, ABC, – DEB= suppt. of BCD= 1 ABC, and DE, EB=AB, BC respectively; .. DB = AC.

Page 73. 1. (1) The A 8 ABD, ACD are equal in area (1. 37). Take away AKD from each and the remainders are equal.

(2) AEAB A ABC= ABCD = ACDF. .. AS EAB, AKB together = A$ CDF, DKC.

2. See solution to Ex. 3 on p. 65.

3. Let AB be the base of given A ABC, and let PQ be the st. line in which the vertex is to lie. Through c draw CD part. to AB meeting PQ in D; then ADB is the required A (1. 37].

4. Let AB be the base of the given A ABC. Through C draw CD par. to AB. Bisect AB in E, and through E draw EF perp. to AB meeting CD in F. Then AFB is the required A (1. 37, and 1. 4].

5. The As are on equal bases and of equal altitude.

6. In the fig. to 1. 34 let the diagonals intersect in E. Then, by Ex. 5 on p. 64, E is the middle pt. of each diagonal.

.. A AEC=AAEB=A BED= ADEC.

7. The ΔΑΒX ΔACX; and ΔYBX AYXC. Hence A ABY= A ACY.

8. Join BD cutting AC in E, then AE is a median of A ABD.

9. Since the equal sides contain supplementary angles the two As can be placed having one side in common and the other two equal sides in same st. line. Thus we have two As of same altitude on equal bases.

10. Let AB, BC be bisected in X, Y respectively; then A BXY = A AXY [1. 38), also A BXY = AXYC; .. ZAXY = A CXY. :. XY is par. to BC (1. 39].

11. Join AD, BC. Then A' AOC, AOD together= A'BOD, AOD. That is, ADAC = A ADB. .: BC is parl. to AD (1. 39).

12.

AAEF = A ABC (1. 38]. .. AAEF = A DEF; :. AD is par to EF (1. 39].

Page 74.
1. Parm. AY = half parm. AC, and AAZB = half parm. AY.
2. Let BD be a diagonal of sq. ABCD.

At B draw BE perp. to DB meeting DC produced in E. Then _ CBE = half a rt. L, and A* DCB, ECB are equal in all respects [1. 26). : . DB = BE, and ADBE = twice A DCB = given square.

3. AS AXB, BYC are each of them half of given parm.

4. Through p draw XY par! to AB or DC. Then AAPB is half parm. Ay, and A DPC is half parm. XC.

Page 75. 1. Let ABCD be the given square. Join BD. Through draw CE par. to BD meeting AD produced in E; then DBCE is a parm. equal to sq. AC and having _ DBC equal to half a rt. L.

2. Let ABCD be the given parm. With centre D and radius DC describe a circle cutting AB in E. Draw CF par'. to De meeting AB produced in F. Then EDCF is a rhombus and it is equal to parm. ABCD. If DC is less than perp. from D to AB the circle will not meet AB and the construction fails.

Page 83.
1. (i) < GBA = alt. Z AHC each being half a rt. L.

(ii) _ FAB = half a rt. 4 = L KAC. i. _ FAB, BAC, CAK together = 2 rt. _ s.

(iii) Let FC meet AB in M and AD in N. Then in A $ FBM; ANM, LBFC = BAD, since As FBC, ABD are equal in all respects, and Z FMB = LAMN (1. 15).

:: LFBA= _ ANM (1. 32].
.. AD is at rt. _ $ to FC.

(iv) Since _ $ FBA, DBC are rt. 28, 1. LFBD is supp. of LABC (1. 15, Cor. 1].

Similarly L KCE is suppt. of L ACB. Hence the result follows by Ex. 9, p. 73.

sq. on

2. Take the case of the exterior squares; then _ CAG = L BAH, each being the sum of LGAH (or – BAС, according as the LA is obtuse or acute) and a rt. L. Also CA, AG = HA, AB... A$ GAC, BAH are equal in all respects [1. 4]. The other case is similar.

3. It is easy to see that AS ACX, BCY are equal in all respects [1. 4]. .. AX = BY. Similarly BY = cz.

4. Let BD be the diagonal of sq. ABCD. Then DB = sum of sqq. on DC, BC= twice the sq. on DC.

5. AX bisects BC (I. 26]; .. sq. on BC = 4 times sq. on BX. That is sq. on AB = 4 times sq. on BX. But sq. on AB = sum of sqq. on BX, AX; .. sq. on AX = three times sq. on BX.

6. Let AB, CD be sides of the two squares ; at B draw BE perp. to AB, and equal to CD. Join AE; then sq. on AE:

= sum of sqq. on AB, BE. 7. Sq. o AB =

= sum of sqq. on AX and BX. Sq. on AC = sum of sqq. on AX and xc.

:. diff. of sqq. on AB and AC = diff. of sqq. on BX and xc. 8. By Ex. 7,

sq. on BZ ~ sq. on AZ sq. on OB ~ sq. on OA. Write down similar results for the other sides, and add.

THEOREMS AND EXAMPLES ON BOOK I.

I. ON THE IDENTICAL EQUALITY OF TRIANGLES.

Page 90.
1. Let the base BC be bisected by the perp. AD.

Then
DC, AD is common and the LBDA = L CDA. .. AB = AC.

BD=

2. Let AD bisect the vert. L. Then - BAD = L DAC; LBDA= LCDA and AD is common. :. AB=AC [1. 26]. 3. Let BC the base be bisected by AD which bisects the vert.

Produce AD to E, making de equal to DA, and join CE. Then AS BAD, EDC are equal [1. 4]. .. AB = EC and Z BAD = L CED i. _ CED= . DAC; .. AC = EC = AB.

L.

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