D. 5 Conc. 6 Rec.

Thus in both cases

AD > C

AF.

Therefore, of alls applied to AB, and deficient s each sim. and similarly situated to AB, AD is the greatest.

by

the

Q. E. D.

Otherwise the Proposition may be enunciated;-Of all the rectangles contained by the segments of a given st. line, the greatest is

the square which is

described on half the line.

SCH. DE CHALES, LARDNER, and some other geometricians recommend that Propositions 27, 28, and 29 should be omitted as unnecessary; but they were frequently employed by the ancient mathematicians, and are required especially for the solution of several problems.

USE AND APP.-In a given ▲ABL to inscribe the greatest parallelogram possible, having an angle, A, in common with the triangle.

Complete the AE of which LB is the diag.
bis. AB in C, AC in K; AL in N and LN in G ;
and through. s K, C, I, draw ||s to AL,

and through. s G, N, P draw Is to AB.

then parallelog. AD, with A com. to ▲ABL, is the greatest parallelog. possible in ABL.

D. 1 24, VI.

2 26, VI.

3

4 C. 2-4.

5 27, VI.

..the defects of parallelogs AO, AD &c. are parallelogs. sim. to parallelog. CM;

... those parallelogs. are about the same diag. BL;

pargs. AO, AD, AF &c. are inscribed in the given ▲ ABL; And. parallelog. AD is described on AC half the base, .. parallelog. AD is the greatest parallelog. in the ▲ABL, and the A is com. to the parallelog. & the ▲.

Q. E. F.

PROP. 28.-Prob.

To a given st. line to apply a parallelogram equal to a given rectilineal figure, and deficient by a parallelogram similar to a given parallelogram; but the given rectilineal figure to which the parallelogram to be applied is to be equal, must not be greater than the parallelogram applied to half of the given line, having its defect similar to the defect of that which is to be applied; that is, to the given parallelogram.

"To given rt. line to apply a parallelogram equal to a given rectilineal figure and deficient by a figure similar to the given parallelogram."-EUCLID.

CON. 10, I. To bis. a given finite st. line, 18, VI. 31, I. 25, VI. 3, L

DEM. 21, VI. 26, VI. Ax. 3. 43, I. 36, I. Ax. 1. 24, VI.

let fig. C be the

Let AB be the given st. line, and C the given fig.
to which the parallelogram is to be equal;
on the half line AE, of
which the defect is sim. to the defect of that
which is to be applied;

3.

and let fig. D be the
be similar.

4 Quæs.

C. 1 10, I.
18, VI.
231, I.

To apply to AB a

to which this defect is to

=

C, and deficient from

sim. to fig. D. EF sim.

the on the line AB, by a
Bis. AB in E; and on EB desc.
and similarly situated to fig. D;
complete the AG, either C, or > C.

CASE I. Suppose AG = fig. C;

=

C.

25, VI.

Make

AG

AG, & ..

fig. C, but > C; and by 36, I, EF also > fig. C.

KM excess of EF above fig. C,

& sim. and simly situated to fig. D.

D. 1C.1.21,VI ... fig D is sim. to

.. line EG > KL, & GF > LM.

53, I. 31, I. Make GX = LK, GO =

XO is sim. to CEF,

7s XO & EF both are about diag. GB.

Let GPB be their diag.; and complete the

scheme.

then EF
ΧΟ

C+ KM, and a part

KM a part of the other;

D.11 Ax. 3. 12 43, I.

13 C. 1,36,I. Ax. 1.

14 Add.

15 D. 11. 16 Rec.

24, VI.

.. rem. gnomon ERO= rem. C.
And.: OR = parlm. XS, .. on adding SR
to each, OB = parlm. XB;

but AE =

EB; .. parlm. XB = & parlm. TE=parlm. OB; to each add parlm. XS;

parlm. TE

fig. C.

.. the whole TS the gnomon ERO; but ERO= fig. C; .. also parlm. TS .. parlm. TS = fig. C is applied to st. line AB, deficient by parlm. SR sim. to fig. D, because SR is sim. to EF.

Q. E. F.

SCH. 1. The Proposition may be thus enunciated; To divide a given st line AB, so that the rectangle contained by the segments may be equal to a given space, as the square on C; but that given space must not be greater than the square of half the given line.

Bis. AB in D.

C.

10, I.

D. 1 Sup. 1.

2 Sup. 2. 3 Sim.

If AD2 = C2 the Problem is solved;

but if AD2C2, & AD > C;

then the Solution though differing in form will be like the foregoing Prop. 28 in substance.

C. 13, I. 11, I. At rt. s to AB draw DE= C ;

2 3, I.

& prod. ED so that EF = AD or DB;

3 Pst. 3 & 1, I from E with rad, EF cut AB in G, and join EG;
4 Sol.
then AB is div. in G so that AG. GB = C2

SCH. 2. Prop. 28. bk. VI. is equivalent to the Problem;-" To inscribe in a given a parlm. equal to a given figure not greater than the maximum inscribed parlm, and having an L in common with the A." Manual of Euclid, Pt. II. p. 98.

The Demonstration is similar to that in Use and App. 27, VI.

USE & APPL. There are two cases of this Proposition which are not unfrequently employed by geometers.

1o. A variation of Sch. 1, 28, VI. To a given st. line AB to apply a rect. AH, deficient by a square GK, which rect. shall be equal to a given square, that on line C; but the given square on C must not be greater than the square on the half AD of line AB.

produce ED so that EF =
AD or DB;

5 Pst. 3 & 1,I from cen. E, with EF desc.

arc meeting AB in G; and
join EG;

6 46, I. 31, I, on GB desc. the square GK, and complete rect. GL; 7 Sol. then ☐ AH = C2, and deficient by GK, has been applied to AB.

D. 1 C.

25, II. 47, I,

3 Sub. Ax. 3.

4 C. 6.

5

6 Conc.

. AB is div. equally in D, and unequally in G;

.. AG.GB +DG2 = DB2 = EF2 = EG2 = ED2 +
DG2.

Take DG2 from each, .. AG. GB = ED2, i. e. = C2.
But AG. GB is

AH C2.

AH,.. GH = GB;

AH C2 has been applied to AB, deficient by the sq. GK.

=

O. E. F.

2°. To a given st. line to apply a rectangle, which shall be equal to a given rectangle, and be deficient by a square; but the given rectangle must not be greater than the square upon half the given line.

Let AB be the given line, and on lines C & D a given

2

not greater than (AB)2;

2

= C. D,

deficient by a square.

On the same side of AB, at
A & B, draw AE, BF each
LAB;

22, I. 10, I. make AEC and BF

3 Pst. 3, I.

=

D;

join EF and bisect it in G; from G, with rad. GE, desc. a meeting AE in H;