ber of equivalent parts by lines drawn from any point in the parallelogram selected at random.

904. It is required to divide a parallelogram into any number of equivalent parts by lines parallel to a given line.

905. It is required to divide a parallelogram into any number of parts, whose areas shall be in a given ratio, by lines parallel to a given line.

It is required to divide a trapezoid into two equivalent parts by a line drawn

906. Parallel to the bases.

907. Perpendicular to the bases.

908. Parallel to one of the legs.

909. Through one of its vertices.

910. Through a given point in one of its bases.

911. Through any point selected at random in its perimeter. 912. Through any point selected at random in the trapezoid. 913. Parallel to a given line.

It is required to divide a trapezoid into any number of equivalent parts by lines drawn

914. Parallel to the bases.

915. Perpendicular to the bases. 916. Parallel to one of its legs.

917. Through one of its vertices.

918. Through any point selected at random in one of its bases.

919. Through any point selected at random in its perimeter. 920. Through any point selected at random in the trapezoid. 921. Parallel to a given line.

It is required to divide a given trapezoid into any number of parts whose areas shall be in a given ratio by lines drawn 922. Parallel to the bases.

923. Perpendicular to the bases.

924. Parallel to one of the legs.

925. Through either vertex.

926. Through any point selected at random in one of the bases.

927. Through any point selected at random in its perimeter. 928. Through any point selected at random in the trapezoid. 929. Parallel to a given line.

930. It is required to divide a trapezium into two equivalent parts by a line drawn from either vertex.

931. It is required to divide a trapezium into two equivalent parts by a line drawn from any point selected at random in its perimeter.

932. Given the diameter of a circle, it is required to construct a straight line equal in length to the circumference. From Theorem 519 it is evident that it can only be approximated.

Let AB be the given diameter and C its middle point. Extend AB indefinitely as AS. Make BD and DE each equal to AB. At E erect the perpendicular EQ, and on it make EF and FG each equal to AB. Join AG, AF, DG, and DF. Lay off EH and HK each equal to AG, and from K lay off

KL equal to AF. Again, from L make LM equal to DG, and MN equal DF. Bisect EN at P; bisect EP at R; and then trisect ER at T and W. Then CT will be the required line approximately equal to the circumference of the circle whose diameter is AB; for, calling the diameter unity,

CE=21,

EL=2CH-KL=2√13-√10,

LM=√5,

MN=√2.

.. EN=2√13−√10+√5+√2, and

ET = √(2 √13 −√10+√5 +√2), and therefore
CT=2}+† (2√13−√10+√5+√2) = 3.1415922 +.

Q.E.F.

APPENDIX.

1. The following theorem, and demonstrations of it, are given as a substitute for 234, for those teachers who may prefer it. Ratio and proportion (261), however, should be taken previous to attempting it.

2. In the same or equal circles two central angles are in the same ratio as the arcs which their sides intercept.

N

QQ

K

Post. Let NBA and QKH be two equal circles, and C and D two central angles.

We are to prove ZC:ZD:: Arc AB: Arc HK.

CASE I. - When the angles are commensurable.

Dem. If the angles are commensurable, there is some angle, as W, which will be contained an exact number of times in each.

Suppose it is contained m times in C, and n times in ≤ D. Then

ZC:LD::m: n.

If, now, lines be drawn from C and D dividing the two angles into m and n equal parts respectively, each part being equal to angle W, then arc AB will be divided into m equal

arcs, and HK into n equal arcs, the divisions all being equal, by Theorem 191.

... Arc AB: Arc HK::m: n.

.. ZC: ZD:: Arc AB: Arc HK. (Theorem 288.)

Post. Let C and D be two incommensurable central angles in equal circles.

We are to prove C:LD:: Arc AB: Arc HK.

Conceive them to be applied as in I.

Then if arc HK is not the fourth term of this proportion, some other arc greater or less than HK must be. Suppose it to be greater as AW.

Then

ZACB: ZACK:: Arc AB: Arc AW.

(a)

Now conceive the arc AB to be divided into equal parts by continued bisection until each part is less than KW. Then there must be at least one point of division between K and W, as N.

.. Z ACB: ZACN:: Arc AB: Arc AN.

.. ZACK: Z ACN :: Arc AW: Arc AN.

(b)

(Theorem 300.)

But the arc AN is less than the arc AW. Consequently, if the proportion be a true one, the angle ACN must be less