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Occasionally it will be found that with certain data in the following exercises the loci do not intersect, or the solution becomes impossible. As in the case given, it will not be difficult to see that the circles would not intersect unless any two of the sides were greater than the third side.


I. Find a point in a given straight line at equal distances from two given points. Construct the figures for all cases.

2. Find a point in a given straight line at a given. distance from a given straight line.

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.


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 of 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

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

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 § 4, Ex. 5); 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 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.




On a given straight line to describe a square.

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.


Divide half a right angle into six equal parts.

II. 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.


By equivalent figures are meant figures whose areas are equal although the figures may be of different shapes, and therefore not conceivable as superposed on one another. Thus a circular field may be as large as a square one, or a triangular piece of paper as large as a rectangular piece, and in such cases these figures would be called equivalent. The consideration of equivalent figures is an important part of Geometry.

Def. 36. The altitude of a parallelogram is the perpendicular distance between one side which is called the base and the side parallel to it.

Thus in the figure at the side the perpendiculars DE, FG, or CH, which are equal (by Theorem 21) since DEGF, DEHC are parallelograms, are each of them the altitude of the parallelogram ABCD, AB being the base.




Any side of a triangle being taken as base, the altitude of the triangle is the perpendicular let fall on that side, or that side produced, from the opposite angle.


Parallelograms on the same base and between the same

parallels will be equivalent.

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