The same diagram and letters will serve for examples 6-10 inclusive. 6. ▲ 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 ▲ ABD, BDC, AB = DC, BD is common, AD = BC; .. ABD = 4 BDC. .. each is a rt. ▲ [1. 29].

S

9. In ▲ ABD, BDC if 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 ▲ OPB, OQC are equal in all respects [1. 26]. .. OP=OQ.

BC =

11. In parms. ABCD, EFGH, AB = EF, FG, and LABC = LEFG. Then ABC may be made to coincide with ▲ EFG [1.4] and AADC will coincide with ▲ EHG [1. 7].

13. Let AP, CQ be perp. to diag. BD. Then ▲ QBC, APD = CQB, and AD = BC. equal in all respects [1. 26].

ADP = alt.

AAPD,

BQC are

14. AX is equal and par1. to YC.

.. &c. [1. 33].

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. ▲ 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 × 2 × 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 spectively. Then XEC, CEA, AEY = 43XEB, spectively. Thus sum of $XEC, CEA, AEY = 2 rt.

BEC, AED re-
BED, DEY re-

8.

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

6. In fig. to Prop. 21, suppose BD, CD bisect base, and let AD be produced to F; then BDF = sum of DBA, BAD ; LCDF = sum of $ DCA, CAD. Hence BDC= the angle at A together with half the sum of the base angles.

7. In AABC let the external bisectors of 4 B and C meet at F, and the internal bisectors at E. Then by Ex. 2, page 29 each of EBF, ECF is a rt. . .. LF is the suppt. of that is F = sum of $ EBC, ECB.

E;

8. Let the st. lines intersect at O, 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 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 par1. to PY.

Then the AOQR, OQX are equal in all respects [1. 26].

.. 4 RQO = 4 XQO. But since RQ is par1. to YP; halves of these are equal; that is, par1. to OQ.

.

RQX =

=

YPX; and the OQX = = L SPX. ... SP is

9. AP=AQ; .. LAPQ = LAQP; .. each is half the suppt. of BAC. Hence APQ = L ABC. .. PQ is par1. to BC [1. 28].

10. Join AD, BE, CF; then AD is equal and par1. to BE, and CF is equal and par1. to BE [1. 33]. .. AD is equal and par1. to CF; .. AC is equal and par1. to DF.

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

4

3

12. In the quadril. ABCD, let AB be par1. to CD, and AD equal to BC. Draw BE par. to AD; then ABED is a parm, and BEAD = BC. .. BCD BED = the suppt. of DAB. Again, since the fours of the fig. are together equal to four rt. 8, the other pair of opp. S are supplementary. Join AC, BD; then in ▲ DEB, ABC, ▲ DEB = suppt. of BCD = ABC, and DE, EB = AB, BC respectively; .. DB = AC.

8

Page 73.

1. (1) The ▲ ABD, ACD are equal in area [1. 37]. Take away AAKD from each and the remainders are equal.

(2) ΔΕΑΒ = AABCA BCD = ACDF. .. AS EAB, AKB together = A CDF, DKC.

2.

8

See solution to Ex. 3 on p. 65.

3. Let AB be the base of given ▲ ABC, and let PQ be the st. line in which the vertex is to lie. Through C draw CD par1. to AB meeting PQ in D; then ADB is the required ▲ [1. 37].

4. Let AB be the base of the given AABC. Through C draw CD par1. to AB. Bisect AB in E, and through E draw EF perp. to AB meeting CD in F. Then AFB is the required ▲ [1. 37, and 1. 4].

5.

8

The ▲ 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.

.. AAEC=^AEB=▲ BED = A DEC.

AACX; and AYBX = AYXC. Hence

8. Join BD cutting AC in E, then AE is a median of ▲ 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 ▲ BXY = ^AXY [1. 38], also ▲ BXY = ^XYC; .'. ^ AXY = ▲ CXY. .. XY is par1. to BC [I. 39].

11. Join AD, BC. Then As AOC, AOD together ▲ BOD, That is, ADAC = ▲ ADB. .. BC is par1. to AD [1. 39].

AOD.

AABC [I. 38]. :. AAEF = A DEF; .. AD is

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 L CBE and ▲ DCB, ECB are equal in all respects [1. 26]. and ▲ DBE=twice ▲ DCB = given square.

=

half a rt. L,

.. DB:

=

BE,

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

4. Through P draw XY par1. to AB or DC. half parm. AY, and DPC is half parTM. XC.

Then ΔΑΡB is

Page 75.

1.

Let ABCD be the given square. Join BD. Through C draw CE par1. to BD meeting AD produced in E; then DBCE is a parm. equal to sq. AC and having

DBC equal to half a rt. 4.

2. Let ABCD be the given parm. With centre D and radius DC describe a circle cutting AB in E. Draw CF par1. 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 (ii) FAB together = 2 rt. ≤ §.

(iii) Let FC meet AB in M and AD in N. Then in ▲ FBM, BFC = BAD, since ▲ FBC, ABD are equal in all respects, FMB = AMN [1. 15].

FBA =

.. AD is at rt.

(iv) Since

ANM [1. 32].

LABC [1. 15, Cor. 1].

S to FC.

FBA, DBC are rt. 4, .. 4 FBD is suppt. of

Hence the result follows

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

3. It is easy to see that ▲ ACX, BCY are equal in all respects .. AXBY. Similarly BY = CZ.

[1.4].

DB

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

Then sq. on

5. AX bisects BC [1. 26]; .. sq. on BC = 4 times sq. on BX. That is sq. on AB = 4 times sq. on BX.

sqq. on BX, AX; .'. sq. on AX

=

But sq. on AB = sum of

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. on 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.
DC, AD is common and the

BD =

2. Let AD bisect the vert. < BDA = CDA and AD is common.

3. Let BC the base be bisected by AD which bisects the vert. L. Produce AD to E, making DE equal to DA, and join CE. Then ▲ BAD, EDC are equal [1. 4]. .. ABEC and ▲ BAD = 2 CED .. CED = 4 DAC ; .'. AC = EC = AB.

S