DEFINITION.

A square is a four-sided figure, having all its sides equal to one another, and all its angles right angles.

PROPOSITION XXVIII.

If two adjacent sides of a rectangle are equal, the figure is a square.

B

Let ABCD be the rectangle contained by AB, AD,
and let AB be = AD.

Then ABCD is a square.

=

For since ABCD is a,.. AD is BC, (1. 26)

and AB is DC, (1. 26)

=

also, by hypothesis, AB is = AD;

.. ABCD is equilateral.

It is also rectangular ;

.. ABCD is a square.

PROPOSITION XXIX.

If the sides and angles of one rectilineal figure are respectively equal to the corresponding sides and angles of another, the figures are equal in all respects.

Let the sides and angles of the rectilineal figure OPQRS be respectively equal to those of CDFGH.

Then shall these figures be equal in all respects.

For let the figure OPQRS be applied to CDFGH, so that may fall on C and OP along CD, then P will coincide with D, and .. PQ will fall on DF, 4 OPQ is = L CDF, and so on. Thus the figure OPQRS will coincide with CDFGH, and will.. be = it in all respects.

COR. If two adjacent sides and the included of one are respectively equal to two adjacent sides and the included L of another, then theses are equal in all respects; therefore also if two adjacent sides of one rectangle are equal respectively to two adjacent sides of another, the rectangles are equal in all respects.

N.B. If a, b are two straight lines, then a rectangle having two of its adjacent sides respectively = a, b will be spoken of as the "rectangle contained by (a, b),” which will be written thus, "rect. (a, b).”

PROPOSITION XXX.

If the sides of a polygon be produced in order, the exterior angles shall together be equal to four right angles.

Let the sides of the polygon NOPQRS be produced in order,

Then shall the exterior angles be together= 4 right 4 s.

Through any point K draw Ku, Kv, Kw, &c. || to OU, PV, QW, &c.

Then the exterior s UOV, VPW, WQX, &c. of the polygon are respectively the s uKv, vKw, wKu, &c.,

=

and are.. together four right ▲ s.

=

(1. 21) (1.9)

COR. All the interior angles of a polygon together with four right angles are equal to twice as many right angles as the figure has sides.

For the NOP of the polygon together with the exterior UOV makes up two right 4 s.

(I. 9)

Hence all the interiors of the polygon together with the adjacent exteriors are equal to twice as many right 4s as the figure has sides;

but all the exteriors are = 4 rights;

.. all the interiors of the polygon together with four rights are = twice as many rights as the figures has sides.

C. G.

4

PROBLEM H.

To divide a straight line into any given number of equal parts.

Let it be required to divide the straight line AB into five equal parts.

Through A draw any other straight line AC:
from it cut off any part AD,

and also DE, EF, FG, GH each equal AD;

join HB, and through D, E, F, G draw lines to HB

and cutting AB in X, Y, Z, W;

then shall AB be divided into five equal parts.

Draw DM, EK || to AB.

Then the 4s of ADME are the 4s of ▲ AXD; (1. 31)

=

and DE is = AD.

Δ

·. DM = AX (1. 3); .'. XY= AX.
Similarly YZ, &c. are each = AX,

and.. AB is divided into five equal parts.

(I. 26)

In like manner may a straight line be divided into any

given number of equal parts.