BOOK I.

having one of its angles CEF equal to the given angle D. Which was to be done.

[In making the angle CEF equal to
D, use the method pointed out in the
Note to Prop. XXIII. Thus:

Mark off DK, DL equal to each other,
and EC equal to either of these, and D
with centre E, and distance EC, describe
the arc F. Then with centre C, and
distance equal to KL, touch the arc F.
Join FE. The angle FEC will be the
required angle.]

K

E

PROP. XLIII. THEOR.

The complements of the parallelograms, which are about the diameter of any parallelogram, are equal to one another.

* xxxiv. 1. the triangle ABC is equal to the triangle ADC: And, because EKHA is a parallelogram,

C

the diameter of which is AK,

the triangle AEK is equal to the triangle AHK:

By the same reason,

the triangle KGC is equal to the triangle KFC: Then, because the triangle AEK is equal to AHK, and the triangle KGC to KFC;

the triangle AEK, together with the triangle KGC, is equal to the triangle AHK,

together with the triangle KFC:

But the whole triangle ABC is equal to the whole ADC; therefore the remaining complement BK

is equal to the remaining complement KD. Wherefore the complements, &c. Q. E. D.

PROP. XLIV. PROB.

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.

Let AB be the given straight line,

and C the given triangle,

and D the given rectilineal angle.

It is required to apply to the straight line AB

a parallelogram equal to the triangle C,

and having an angle equal to D.

BOOK I

Make the parallelogram BEFG equal to the triangle C, and having the angle EBG equal to the angle D,

. xlii. 1.

BOOK I. so that BE be in the same straight line with AB, and produce FG to H;

xxxi. 1. and through A draw AH parallel to BG or EF, and join HB.

Then because the straight line HF

falls upon the parallels AH, EF,

с

• xxix. 1. the angles AHF, HFE are equal to two right angles;

d 12 Ax.

wherefore BHF, HFE are less than two right angles:

But straight lines which with another straight line
make the interior angles upon the same side

less than two right angles,

do meet if produced far enough:

Therefore HB, FE shall meet if produced;

let them meet in K,

and through K draw KL, parallel to EA or FH,
and produce HA, GB to the points L, M:
Then HLKF is a parallelogram,

of which the diameter is HK;

and AG, ME are the parallelograms about HK;
and LB, BF are the complements;

• xliii. 1. therefore LB is equal to BF:

f xv. 1.

But BF is equal to the triangle C;

wherefore LB is equal to the triangle C ;

and because the angle GBE is equal to the angle ABM, and likewise to the angle D;

the angle ABM is equal to the angle D:

Therefore the parallelogram LB is applied to the straight

line AB,

is equal to the triangle C,

and has the angle ABM equal to the angle D.

Which was to be done.

[In order to draw this figure correctly, a triangle must first be placed upon the line AB produced, equal to C. Thus:

BOOK I.

E

AV

с

Measure BO equal to one of the sides of C, and, as in the
Note to Prop. XXIII., make the other sides OP and BP
equal respectively to the two other sides of the triangle.
Then proceed with the triangle PBO just as the figure in
Prop. XLII. was drawn.]

BOOK I.

a xlii. 1.

b

PROP. XLV. PROB.

To describe a parallelogram equal to a given rectilineal figure, and having an angle equal to a given rectilineal angle.

Let ABCD be the given rectilineal figure,

and E the given rectilineal angle.

It is required to describe a parallelogram equal to ABCD, and having an angle equal to E.

Join DB,

and describe the parallelogram FH

equal to the triangle ADB,

and having the angle HKF equal to the angle E,

and to the straight line GH

xliv. 1. apply the parallelogram GM equal to the triangle DBC, having the angle GHM equal to the angle E;

c xxix. 1.

and because the angle E is equal to each of the angles FKH, GHM,

the angle FKH is equal to GHM:

add to each of these the angle KHG;

therefore the angles FKH, KHG are equal to the angles

d xiv. 1.

and because at the point I in the straight line GH,

the two straight lines KH, HM,

upon the opposite sides of it,

make the adjacent angles equal to two right angles,

KH is in the same straight lined with HM,

and because the straight line HG