47.

Let AB be the st. line to be divided, K a side of the
At B make ABD equal to half a rt. .

From centre
From C (or

given sq. A with radius K draw a circle to cut BD at C and C'. C') draw CX perp. to AB. Then AB is divided as required at X. For sq. on AC=sqq. on AX, XC [1. 47] =sqq. on AX, XB,

for XC = XB, since CXB is a rt. 4, and XBC is half a rt. ▲ [1. 32, and 1. 6].

There will be two solutions, one solution, or no solution, according as the circle with radius K cuts BC in two pts., touches at one, or does not meet it at all.

IX. ON LOCI. Page 117.

4. The locus is a concentric circle whose radius is equal to the sum or difference of the radius of given circle and the given distance.

5. Let OA, OB be the two intersecting st. lines, K the given const. length. At O draw OC perp. to OA and equal to K; draw CD par1. to OA meeting OB in D. In OA make OE equal to OD and join DE. Then DE is the required locus; for by Ex. 22, p. 99, the sum of the perps. on OA and OB from any pt. in DE is equal to the perp. from D on OA = OC = K.

6. With the same lettering and construction as in Ex. 5, let DE be produced both ways indefinitely to X and Y. Then the required locus is the part of XY external to the AODE. [See Ex. 23, p. 99.]

7. Let AB be the rod of given length, and C the intersection of the rulers at right angles. Bisect AB in D; then CD = half of AB. Therefore the locus of D is a circle whose centre is the fixed pt. C and whose radius is half the given length of the rod.

8. Let ACB be a rt. angled ▲ on AB as hypotenuse. Then, if AB is bisected at D, CD is constant, being equal to half of AB. Thus the locus is a circle whose centre is D and radius equal to half the given base.

9. Bisect AB in C and AX in D and join DC. Then the locus required is the locus of the vertices of right angled ▲ on base AC as hypotenuse [Ex. 8].

10. Let AB be the base, and AD the altitude, which is known since the base and area are given. Then the locus required is a st. line through D par'. to AB [1. 39].

11. The diagonals of a parm. divide it into four equal having their vertices at the inters". of the diagonals. Thus the problem is the same as in Ex. 10, and the required locus is a st. line through the inters". of the diagonals par1. to the given base.

12. Let BC be the given base, and ABC the ▲ in any of its positions. Then since the area of the ▲ is constant, A must move on a fixed st. line par1. to BC. Let AX be the median bisecting BC; then if o is the inters". of medians AO = 20X [Ex. 4, p. 105]. Draw AD perp. to BC and in it take AP = 2PD [Ex. 19, p. 99]. Join OP. Then it may be shewn [as in Ex. 2, p. 96] that OP is par1. to BC. But P is clearly at a fixed distance from BC (being one-third of the distance between the parl.); .. O lies on a st. line par1. to BC and at a fixed distance from it.

X. INTERSECTION OF LOCI. Page 119.

Join

1. Let X, Y be the given pts., AB the given st. line. XY and bisect it in C; draw CZ perp. to XY meeting AB in D. Then from equal ▲ DCX, DCY [1. 4] DX = DY.,

2. If the lines are not par1. let them be AB, CB meeting in B; and let P, Q be the given distances. Draw BE perp. to AB and equal to P; at E draw EF par1. to BA; then the required pt. lies in EF.

Again draw BD perp. to BC and equal to Q; at D draw DF par. to BC; then the required pt. lies in DF. That is, F is the required pt.

There are four solutions since each of the lines BE and BD may be drawn in either direction.

3. Let AB be the given base, X the given 4, and Y the length of the side opposite. At A draw AZ making BAZ equal to X. From B as centre and with radius equal to Y describe a circle. Draw BD perp. to AZ; then BD is the shortest distance of B from AZ.

(i) If Y< BD, the circle will not cut AZ, and there is no possible with the given parts.

(ii) If Y = BD, the circle will meet AZ in one pt., and there is one solution, viz. the right-angled ▲ BAD.

(iii) If Y> BD, the circle will cut AZ in two pts. C1, C, and there will be two triangles ABC1, ABC, each of which satisfy the data. This last case however requires that Y shall be less than AB, for then both pts. C1, C, will lie on AZ on the same side of A. If one of these pts. C2 lies on opposite side, the ▲ BAC, would have the L BAC equal to the supplement of given X, and would not satisfy the data.

2

4. Let ABC be the given ▲ on base BC, and DE the given st. line. Through A draw AF par1. to BC meeting DE in F and join FB, BC. Then FBC is the required ▲ [1. 37]. If ED is par1. to BC the solution is only possible when DE passes through A. In this case any pt. in DE may be taken as the vertex of the required A, and the number of solutions is unlimited.

5. Let ABC be the given ▲ on base BC. Draw AD par1. to Bisect BC at E and draw ED perp. to BC meeting AD in D. Then BDC is the required A [1. 4 and I. 37].

BC.

6. Let AB, CD be the two given par1. st. lines, and X the given pt. Draw AC perp. to CD and bisect it at E. Through E draw EF par1. to either of the given st. lines. Then the required pt. must lie in this st. line. With centre X and radius equal to the given distance describe a circle cutting EF in P and Q; then P and Q satisfy the required conditions. If the circle meets EF in one pt. only there is one solution; if the circle does not meet EF there is no solution.

BOOK II.

EXERCISES.

Page 129.

(i) AP = |AB = { {AQ + QB}.

(ii) Mark off AQ' equal to BQ. Then P is clearly the middle point of QQ', so that

PQ = {QQ' = }} (AQ — AQ') = 1⁄2) (AQ — BQ).

Page 131.

Take the enunciations as given on p. 131, that is,

(i) The two ▲ FAB, HAC are identically equal [1. 4].

..the ABF the LACH,

=

and the ▲ LHB = the vert. opp. AHC;

.. the HBL, LHB the S ACH, AHC,

=

.. the remaining 4 HLB the remaining 4 HAC [1. 32]

=

= a rt. angle.

H. K. E.

3

(ii) Since EF = EB, .'. L EBF = L EFB.

8

And in the ▲ OBL, CFL, we have the 3 OLB, CLF rt. angles,

Hence it may be proved that AOC is a rt. angle after the method of III. 31.

(iii) Produce FG, DB to meet at M.

Then since the sq. FH

the rect. HD, these are complementary parms., and hence H lies on the diagonal CM. [1. 43. Converse.]

Hence it may easily be shewn that the diagonals GB, FD, AK of the rectangles GHBM, FCDM, ACKH are par1.

1.

Page 142.

AB. CD + BC. AD AB. CD + AB. BC + BC2

2. Let ABC be an isosceles ▲, having its vertex at A, and let BD be drawn perp. to AC.

Then each of the

ABC, ACB must be acute. [1. 17.]
Hence by 11. 13, AB2 = AC2 + BC2 — 2AC. CD.

But AB2 = AC2, so that BC2 = 2AC. CD.

3. Let ABC be the ▲, having the

ABC equal to one-third
First prove

of two rt. angles. From A draw AX perp. to BC.

AB = 2. BX: this may be done by joining X to the middle point of AB. [See Ex. 4, p. 59.]

Then by II. 13, AC2 = AB2 + BC2 – 2BX. BC