:: the square of CD,

to a fourth term.

Multiply this latter fourth term by that number of the ratio which corresponds to the part to be adjacent to AB, and the other fourth term by the other number of the ratio; add the two products together, and divide the sum by the sum of the numbers expressing the ratio. The square root of the quotient will give the length of the division line FE. Then,

As sin. E. sin. B:: AB: ES.

The difference between FE and ES, gives FS. Then, As sin. P. sin. E:: FS: AF.*

:

EXAMPLES.

1. Let the bearing of AB be North, dist. 12 ch.; BC N. 56° E. 20.78 ch.; CD, S. 334° E. 22.21 ch.; and DA, S. 80° W. 30 ch.; it is required to divide the tract into two parts by a line FE, running N. 20° W. so that the part ABEF may be to the part FECD as 3 to 5.

Angle A=80°, B=1234°, C=90°, D=66°, E=761°, F=794° and P=24°.

The truth of this rule is evident from the demonstration to the rule in the last problem.

Construction. Draw CG according to the reverse bearing of FE, and then proceed with the construction exactly as in the last problem.

2. The boundaries of a field ABCD are given as follow: viz. AB, S. 10° W. 7.20 ch.; BC, S. 67° W. 12.47 ch.: CD, N. 23° W. 13.33 ch.; and DA, S. 89° E. 18 ch.; and it is to be divided into two parts ABEF and FECD, in the ratio of 3 to 4, by a line FE, running due South. Required the length of the division line FE and the

PROBLEM XIX.

The boundaries of a tract of land ABCDEFGHIA, Fig. 108, being given, to divide it into two equal parts by a line IN running from the corner I, and falling on the opposite side CD.

RULE.

Suppose lines drawn from I, to C and D, and calculate the area of the whole tract.

Take the corrected latitudes and departures* of IA, AB, and BC, and by balancing find the latitude and departure of CI; also calculate the area of the part IABCI; from half the area of the whole tract, subtract the area of the part IABCI, the remainder will be the area of the triangle ICNI.

Take the latitudes and departures of IC and CD, and by balancing find the latitude and departure of DI, and calculate the area of the triangle ICDI. Then,

As the area of the triangle ICDI,

Is to the area of the triangle ICNI;
So is the latitude of CD,

To the latitude of CN.

Also, As the area of the triangle ICDI,
Is to the area of the triangle ICNI;
So is the departure of CD,

To the departure of CN.

* It is the corrected latitudes and departures that are to be used throughout the calculation.

Now take the latitudes and departures of IC and CN, and by balancing find the latitude and departure of the division line NI; with which, find its bearing and distance.*

EXAMPLES.

1. Let the bearing of AB be N. 19° E. dist. 27 ch.; BC, S. 77° E. 22.75 ch.; CD, S. 27° E. 28.75 ch.; DE, S. 52° W. 14.50 ch.; EF, S. 15° E. 19 ch.; FG, West, 17.72 ch.; GH, N. 36° W. 11.75 ch.; HI, North, 16.07 ch., and IA, N. 62° W. 14.88 ch.; it is required to divide the tract into two equal parts by a line IN running from the corner I, and falling on the opposite side CD.

First calculate the whole area, thus:

*This rule needs no demonstration.