Wherefore a ▲ FCG has been described, which Om ABCD, and having the FCG =

gn. 4 E.-Q. E. F.

PROP. XLIII. THEOR.

m, of which

GEN. ENUN.-The complements of the parallelograms, which are about the diameter of any parallelogram, are equal to one another. PART. ENUN.-Let ABCD be a the diam. is AC; EH, FG the ms about AC, i.e. through which AC passes; and BK, KD the other ms which make up the whole figure ABCD, and B

E

A

H

D

K

F

... called complements. Then the complement BK complement KD.

=

DEMONST.-1. Because the diam. AC bisects them ABCD (Prop. XXXIV.), .'. the ▲ ABC

= ▲ ADC.

2. For the same reason, the ▲ AEK = Δ AHK, and the ▲ KGC = ▲ KFC.

3. .. the complement BK = ▲ ABC

(▲ AHK + ▲

Wherefore the complements &c.-Q. E. D.

There are other cases of this Proposition; since it is not necessary that the ms about AC should be united at K, or that the complements should be ms. The Theorem may be equally proved, when thems EH, FG are not connected

Fig. 1.

A

H

D

To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.

PART. ENUN.Let AB be the gn. st. line,

c the gn. A,

D the gn.

rectilineal; then it

is required to apply to

the st. line AB a□m = ▲ с, and having an

L = 4 D.

m BEFG

CONST. and DEMONST.-Make the = ▲ c, and having the EBG = ≤ D (Prop. XLII.), and let BE be in the same st. line with AB; through a draw AH || to BG or EF (Prop. XXXI.), and meeting FG produced in H; join HB.

2. And because the st. line нF falls upon the st. line AH, EF, .. the S AHF+ HFE = 2 rt.s (Prop. XXIX.), ... the BHF + HFE are 2 rt. s; HB, FE will meet if produced towards в and E. (Prop. XXIX. Cor.) 3. Let them meet in K; through K draw KL

B

..

to EA or FH (Prop. XXXI.), and meeting HA produced in L; also produce GB to M.

4. Then HLKF is am, of which the diam. is HK; AG, ME are the ms about HK, and LB, BF are the complements: ... the complement LB = complément BF (Prop. III.) = ▲ C. (Const.)

Δ

5. Also the ABM = vertical / EBG (Prop. XV.) D. (Const.)

=

Wherefore them LB has been applied to the gn. st. line AB, whieh = the gn. ▲ c, and having the ABM = gn. ≤ D.—Q. E. D. The converse Problem may be thus effected.

GEN. ENUN.-To a given straight line to apply a triangle, which shall be equal to a given parallelogram, and having an angle equal to a given rectilineal angle.

PART. ENUN.-Let AB be the gn. st. line, CDEF the gn. m, G the gn. rectilineal . Then it is required to apply to AB a ▲ having an ▲ = L G.

m

the ABKL

=

=

CDEF, and

CONST.-Draw the diam. CF; produce EF to H, making FH = EF, and join CH. To the st. line AB apply A CEH, and having the ABK = 4 G (Prop. XLIV.); produce BK to м, making км = BK, and join AM, AK; then ▲ ABM is E the ▲ required.

K

M

H

DMONST.-For the A ABM = 2 A ABK (Prop. NN.)

= BL (Prop. XLI.) (Prop. NN.) Wherefore to the

ABM, which

G.-Q. E. F.

gn.

=

A CEH (Const.)

CEFD. (Prop. XLI.)

gn. st. line AB has been applied a ▲ m ED, and has the ABM = gn.

PROP. XLV. PROB.

GEN. ENUN.-To describe a parallelogram equal to a given rectilineal figure, and having an angle equal to a given rectilineal angle. PART. ENUN.-Let ABCD be the gn. rectilineal figure, E the

gn. rectilineal; then

it is required to describe a Om ABCD, and having an L = LE.

A

F

E

D

▲ DBC, and having ≤ GHM = ≤ E (Prop. XLIV.); then the figure FKML is the m required.

side the

DEMONST.-1. Since the FKH = GHM (Const.), add to each .. the S FKH + KHG = / (Ax. 2).

S

E =

KHG,

KHG + GHM

2. But the S FKH + KHG 2 rt. <s (Prop. XXIX.); .. the s KHG + GHM = 2 rt. s (Ax. 1).

3. And because at the pt. H in the st. line GH, the two st. lines KH, Hм upon the opposite side of it, makes the adjacent s = 2 rt. ; .. KH is in the same st. line with Hм. (Prop. XIV.)

4. Again, because the st. line GH meets the I st. lines KM, FG, .. the alternate

=

alternate HGF. (Prop. XXIX.)

MHG

5. To each add the HGL, .. the S MHG + HGL = ≤$ HGF + HGL (Ax. 2).

6. But the <S MHG + HGL = 2 rt. s (Prop. XXIX.), .'. the S HGF + HGL = 2 rt. (Ax. 1), and ... FG is in the same st. line with GL. (Prop. XIV.)

7. Now, because KF is || to HG, and HG to ML, .. KF is || to ML (Prop. XXX.); and Kм is also to FL; .. KFLM is a m.

8. Hence the figure ABCD

= ▲ ADB +

FH + om HL (Const.)

=

Wherefore a □m has been described = to the gn. figure ABCD, and having the FKM =gn. E.-Q. E. F.

an

COR. From this it is manifest how to a given st. line to apply a m, which shall have = a gn. rectilineal ▲, and be = a_gn. rectilineal figure: viz. by applying to the st. line am to the first ▲ ABD, and having = the gn. 4.

an

It is also manifest how to proceed, whatever be the number of sides to the gn. rectilineal figure. For it may be divided into as many ▲3, except two, as the figure has sides; and am to each of them in succession is to be applied to the opposite side of each successivem.