11. If the three points of bisection of the three sides of BOOK I. a triangle be joined by straight lines, the triangle so formed shall have its angles equal to those of the original triangle.

Prove, as in the last, the

sides to be respectively par

F

allel to the sides of the triangle, and then use I. 34.

E

B

12. From a given point to draw a straight line, which shall make with a given straight line an angle equal to a given rectilineal angle.

Use in the construction I. 31. and I. 23.; and in the proof, I. 29.

BOOK I.

15. The straight line which bisects the ex-
terior angle at the vertex of an isosceles
triangle, is parallel to the base.

E

17. From a given point in the side of a triangle, to draw

a straight line which shall

divide it into two equal

parts.

Let P be the point.

Bisect AB in O.

Join PO and CO.

Draw CQ parallel to PO.

Join PQ, which shall be the required line.

Show that the triangles POQ and CPO are equal.

B

Add to each APO. Then ACO equals APQ, which, therefore, is equal to half the original triangle.

18. To bisect a parallelogram by a straight line drawn from a given point in one of its B

sides.

Cut off from AB, AQ equal

t oCP.

The straight line joining P

Q

A

D

P

and Q divides the parallelogram as required.

The proof of this is easy, and the learner may find it out BOOK I. for himself.

19. The square described

upon the diagonal of a

given square, is double

of that square.

Prove the four triangles

into which

which the larger F square is divided by joining B and its angular points, to be each equal to either of the two triangles

B

into which the original square is divided.

E

20. If four points be taken in the four sides of a square, at

equal distances from the angular A

points respectively, the straight line

H

E

B

which joins them will form a square.

Prove HEF to be a right angle, and then two adjacent sides of the figure, as HE, FE, to be equal.

F

D

G

BOOK II.

BOOK II.

DEFINITIONS.

I.

Every right angled parallelogram is said to be contained by any two of the straight lines which contain one of the right angles.

II.

E

D

In every parallelogram, any of the parallelograms about a
diameter, together with the A
two complements, is called a
Gnomon. "Thus the paral-
lelogram HG, together with
the complements AF, FC, is

Η

K

ters AGK, or EHC, which are at the opposite angles of the parallelograms which make the gnomon.”

[Observe that the parallelogram EK, together with the complements AF and FC is also a gnomon, and would be expressed by the letters

A

AKG or CEH. In fact, a
gnomon is the whole paral-
lelogram, with the exception
of one of the parallelograms H
about the diameter.]

F

D

THE ELEMENTS OF EUCLID.

87

BOOK II.

PROP. I. THEOR.

If there be two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the rectangles contained by the undivided line, and the several parts of the divided line.

Let A and BC be two straight lines;

and let BC be divided into any parts in the points D, E; the rectangle contained by the straight lines A, BC

and through G draw GH parallel to BC;

с

A

and through D, E, C, draw DK, EL, CH, parallel to BG;

Then the rectangle BH is equal to

the rectangles BK, DL, EH ;
and BH is contained by A, BC,
for it is contained by GB, BC,
and GB is equal to A;
and BK is contained by A, BD,
for it is contained by GB, BD,
of which GB is equal to A;

and DL is contained by A, DE,

because DK, that isd BG, is equal to A ;

and in like manner the rectangle EH is contained by A, EC:
Therefore the rectangle contained by A, BC is equal to
the several rectangles contained by A, BD, and by A, DE;
and also by A, EC.

Wherefore, if there be two straight lines, &c. Q. E. D.

b iii. 1.

C xxxi. 1.

d xxxiv. 1.