3. Find a point in a given straight line at equal distances from two other straight lines.

4. On a given straight line to describe an equilateral triangle.

5.

Describe an isosceles triangle on a given base, each of whose sides shall be double of the base.

6. Find a point at a given distance from a given point, and at the same distance from a given straight line.

7. Given base, sum of sides, and one of the angles at the base, construct the triangle.

8. Given base, difference of sides, and one of the angles at the base, construct the triangle.

9. Find a point at a given distance from the circumference of two given circles, the distances being measured along their radii or their radii produced.

10. A straight railway passes within a mile of a town. A place is described as four miles from the town, and half a mile from the railway. How many places satisfy the conditions ?

II. Find a point equidistant from three given straight lines that intersect one another but not in the same point.

ANALYSIS AND SYNTHESIS.

of

If problems cannot be solved by this method, it remains to attack them by the method, as it is called, of Analysis and Synthesis. This is not so much a method as a way searching for a suggestion, and nothing but experience and ingenuity will here avail the student. The solution is

supposed to be effected, and relations among the parts of the figure are then traced until some relation is discovered which can give a clue to the construction. Nothing but seeing examples can make this clear.

(1) It is required to draw a line to pass through a given point and make equal angles with two given intersecting lines.

Let O be the given point, AB,

AC the given lines.

We reason as follows (analysis): suppose POQ were the line required, then the angle at P= angle at Q.

Therefore AP=AQ; therefore if

we bisected the angle A, POQ would be at right angles to the bisector.

Now this is a suggestion we can work backwards from, and the construction is as follows.

Synthesis. Bisect the angle BAC, and let fall OH a perpendicular to the bisector, and let it meet the lines in P, Q, and POQ can then be proved to be the line required.

(2) It is required to draw from a given point three straight lines of given lengths, so that their extremities may be in the same straight line, and intercept equal distances on that line.

Analysis. Suppose OA, OB, OC were the three lines, so that CBA is a straight line, and CB =BA.

Then it occurs to us that if

OB were prolonged to D, making BD=OB, then CD and DA would be respectively parallel and equal to OA and OC (see § 3, Ex. 21); and that the sides of the triangle DOA are respectively equal to OA, OC and 20B. Hence the construction is suggested.

Synthesis. Make a triangle DOA whose sides are OA, OC, and 20B; complete the figure, by drawing DC, OC parallel to OA, AD; and the other diagonal ABC will be the line required. For it may be shewn that AB = BC.

The student must not be surprised if he finds problems of this class difficult. For there is nothing except previous knowledge of geometrical facts to point out which of the many relations of the parts of the figure are to be followed up in order to arrive at the particular relation which suggests the construction. It is not easy to see what is to suggest the producing of OB to D as in the figure.

Subjoined are a few problems of no great difficulty, which may be solved by this method.

I.

PROBLEMS.

On a given straight line to describe a square.

2. Describe a rectangle with given sides.

3. Given two sides of a parallelogram and the included angle, construct the parallelogram.

4. Given the lengths of the two diagonals of a rhombus, construct it.

5. From a given point without a given straight line. to draw a line making an angle with the line equal to a given angle.

6. Describe a square on a given straight line as diagonal.

7.

Draw through a given point, between two straight lines not parallel, a straight line which shall be bisected in that point.

8. Place a line of given length between two intersecting lines so as to be parallel to another given line.

9. Trisect a right angle.

IO.

II.

Divide half a right angle into six equal parts.

Three straight lines meet in a point, draw a straight line such that the parts of it intercepted by the three lines shall be equal to one another.

12. Trisect a given straight line.

BOOK II.*

EQUALITY OF AREAS.

SECTION I.

THEOREMS.

IN Book I. Theorems 5, 7, 15, 17, 20 we have had instances of figures whose areas are equal, and whose areas are proved to be equal, by shewing that the figures could be placed so as to coincide with one another, or are congruent, or identically equal. But figures of different shapes may nevertheless be equal in area, though they cannot be placed so as to coincide with one another; thus a circular field may be as large as a square one, and a triangle as large as a rectangle.

In the present section we proceed to the consideration of rectilineal figures whose areas are equal, though the figures are not of the same shape.

Def. 1. The altitude of a parallelogram with reference to a given side as base is the perpendicular distance between the base and the opposite side.

Thus in the figure the perpendiculars DE, FG, or CH, which are equal (by I. 28) since DEFG, DEHC are parallelograms, are each of them the altitude of the parallelogram ABCD, AB being the base.

D F

A E

B H

* Book III. (with the exception of its last Section) is independent of Book II., and may be studied immediately after Book I.