measured by the arc AB; therefore, the angle FAB=BCA; but the angle CBA or FBA, is common to both triangles; therefore, the third angle, CAB, of the one triangle, is equal to the third angle, AFB, of the other (th. 11, b. 1, cor. 2), and the two triangles are equiangular and similar.

But the CBA is isosceles; therefore, the ▲ AFB is also isosceles, and AB=AF, and we have the following proportions: CA: AB::AB: BF

Now let AE=c, AB=x, CA=1. Then _AF=x, and EF=c—x, and the proportion becomes,

Also,

1:x:x : BF. Hence BF=x2
FG=2-x2

As AE and GB are two chords that intersect each other at the point F, we have,

GFXFB=AF×FE

(th. 17, b. 3)

If we suppose the arc AF to be 60 degrees, then c=1, and the equation becomes x3-3x=-1; a cubic equation, easily resolved by Horner's method (Robinson's Algebra, University Edition, Art. 193), giving x=.347296+, the chord of 20°. This again may be taken for the value of c, and a second solution will give the chord of 6° 40′, and so on, trisecting as many times as we please.

If the pupil has carefully studied the foregoing principles, he has the foundation of all geometrical knowledge; but to acquire independence and confidence, it is necessary to receive such discipline of mind as the following exercises furnish.

Some of the examples are mere problems, some are theorems, and some a combination of both. Care has been taken in their selection, that they should be appropriate; not very severe, not such as to try the powers of a professed geometrician, nor such as would be too trifling to engage serious attention.

EXERCISES IN GEOMETRICAL INVESTIGATION. 1. From two given points, to draw two equal straight lines, which shall meet in the same point, in a line given in position.

2. From two given points on the same side of a line, given in position to draw two lines which shall meet in that line, and make equal angles with it.

3. If from a point without a circle, two straight lines be drawn to

the concave part of the circumference, making equal angles with the line joining the same point and the center, the parts of these lines which are intercepted within the circle, are equal.

4. If a circle be described on the radius of another circle, any straight line drawn from the point where they meet, to the outer circumference, is bisected by the interior one.

5. From two given points on the same side of a line given in position, to draw two straight lines which shall contain a given angle, and be terminated in that line.

6. If, from any point without a circle, lines be drawn touching it, the angle contained by the tangents is double the angle contained by the line joining the points of contact, and the diameter drawn through one of them.

7. If, from any two points in the circumference of a circle, there be drawn two straight lines to a point, in a tangent, to that circle, they will make the greatest angle when drawn to the point of contact.

8. From a given point within a given circle, to draw a straight line which shall make, with the circumference, an angle, less than any angle made by any other line drawn from that point.

9. If two circles cut each other, the greatest line that can be drawn through the point of intersection, is that which is parallel to the line joining their centers.

10. If, from any point within an equilateral triangle, perpendiculars be drawn to the sides, they are, together, equal to a perpendicular drawn from any of the angles to the opposite side.

11. If the points of bisection of the sides of a given triangle be joined, the triangle, so formed, will be one-fourth of the given triangle.

12. The difference of the angles at the base of any triangle, is double the angle contained by a line drawn from the vertex perpendicular to the base, and another bisecting the angle at the vertex.

13. If, from the three angles of a triangle, lines be drawn to the points of bisection of the opposite sides, these lines intersect each other in the same point.

14. The three straight lines which bisect the three angles of a triangle, meet in the same point.

15. The two triangles, formed by drawing straight lines from any point within a parallelogram to the extremities of two opposite sides, are, together, half the parallelogram.

16. The figure formed by joining the points of bisection of the sides of a trapezium, is a parallelogram.

17. If squares be described on three sides of a right angled triangle,

and the extremities of the adjacent sides be joined, the triangles so formed, are equal to the given triangle, and to each other.

18. If squares be described on the hypotenuse and sides of a right angled triangle, and the extremities of the sides of the former, and the adjacent sides of the others, be joined, the sum of the squares of the lines joining them, will be equal to five times the square of the hypotenuse.

19. The vertical angle of an oblique-angled triangle, inscribed in a circle, is greater or less than a right angle, by the angle contained between the base, and the diameter drawn from the extremity of

the base.

20. If the base of any triangle be bisected by the diameter of its circumscribing circle, and, from the extremity of that diameter, a perpendicular be let fall upon the longer side, it will divide that side into segments, one of which will be equal to half the sum, and the other to half the difference of the sides.

21. A straight line drawn from the vertex of an equilateral triangle, inscribed in a circle, to any point in the opposite circumference, is equal to the two lines together, which are drawn from the extremities of the base to the same point.

22. The straight line bisecting any angle of a triangle inscribed in a given circle, cuts the circumference in a point, which is equidistant from the extremities of the sides opposite to the bisected angle, and from the center of a circle inscribed in the triangle.

23. If, from the center of a circle, a line be drawn to any point in the chord of an arc, the square of that line, together with the rectangle contained by the segments of the chord, will be equal to the square described on the radius.

24. If two points be taken in the diameter of a circle, equidistant from the center, the sum of the squares of the two lines drawn from these points to any point in the circumference, will be always the same.

25. If, on the diameter of a semicircle, two equal circles be described, and in the space included by the three circumferences, a circle be inscribed, its diameter will be the diameter of either of the equal circles.

26. If a perpendicular be drawn from the vertical angle of any triangle to the base, the difference of the squares of the sides is equal to the difference of the squares of the segments of the base.

27. The square described on the side of an equilateral triangle, is equal to three times the square of the radius of the circumscribing circle.

28. The sum of the sides of an isosceles triangle, is less than the sum of any other triangle on the same base and between the same parallels.

29. In any triangle, given one angle, a side adjacent to the given angle, and the difference of the other two sides, to construct the triangle.

30. In any triangle, given the base, the sum of the other two sides, and the angle opposite the base, to construct the triangle.

31. In any triangle, given the base, the angle opposite to the base, and the difference of the other two sides, to construct the triangle.

PROBLEMS REQUIRING THE AID OF ALGEBRA FOR THEIR SOLUTION.

No definite rules can be given for the solution or construction of the following problems; and the pupil can have no other resources than his own natural tact, and the application of his analytical and geometrical knowledge thus far obtained; and if that knowledge is sound and practical, the pupil will have but little difficulty; but if his geometrical acquirements are superficial and fragmentary, the difficulties may be insurmountable: hence, the ease or the difficulty which we experience in resolving such problems, is the test of an efficient or inefficient knowledge of theoretical geometry.

When a problem is proposed requiring the aid of Algebra, draw the figure representing the several parts, both known and unknown. Represent the known parts by the first letters of the alphabet, and the unknown and required parts by the final letters, &c.; and use whatever truths or conditions are available to obtain a sufficient number of equations, and the solution of such equations will give the unknown and required parts the same as in common Algebra.

But as we are unable to teach by more general precept, we give the solutions of a few examples, as a guide to the student.

The first two are specimens of the most simple and easy; the last two or three are specimens of the most difficult and complex, or such as might not be readily resolved, in case solutions were not given.

It might be proper to observe that different persons might draw different figures to the more complex problems, and make different equations and give different solutions; but the best solutions are always the most simple.

PROBLEM 1.

Given, the hypotenuse, and the sum of the other two sides of a right angled triangle, to determine the triangle.

Let ABC be the A. Put CB=y, AB=x, AC=h, and CB+AB=s. Then, by a given condition we we have,

C

A I B

N. B. In place of putting x to represent one side, and y the other, we might put (x+y) to represent the greater side, and (x-y) the lesser

side; then,

h2

Given, the base and perpendicular of a triangle, to find the side of its

That is, the side of the inscribed square is equal to the product of the base and altitude, divided by their sum.

In a triangle, having given the sides about the vertical angle, and the ine bisecting that angle and terminating in the base, to find the base.

Let ABC be the A, and let a circle be circumscribed about it. Divide the arc AEB into two equal parts at the point E, and join EC. This line bisects the vertical angle (th. 9, b. 3, scholium). Join BE.

Put AD=x, DB=y, AC=a, CB=b, CD=c, and DE=w. The two As, ADC and EBC, are equiangular; from which we have,

w+c:ba: c; or, cw+c=ab

(1)