NOTE 9. On the Methods employed in the solution of
Problems.

In the solution of Geometrical Exercises, certain methods may be applied with success to particular classes of questions. We propose to make a few remarks on these methods, so far as they are applicable to the first two books of Euclid's Elements.

The Method of Synthesis.

In the Exercises, attached to the Propositions in the preceding pages, the construction of the diagram, necessary for the solution of each question, has usually been fully described, or sufficiently suggested.

The student has in most cases been required simply to apply the geometrical fact, proved in the Proposition preceding the exercise, in order to arrive at the conclusion demanded in the question.

This way of proceeding is called Synthesis (ovveσis=composition), because in it we proceed by a regular chain of reasoning from what is given to what is sought. This being the method employed by Euclid throughout the Elements, we have no need to exemplify it here.

The Method of Analysis.

The solution of many Problems is rendered more easy by supposing the problem solved and the diagram constructed. It is then often possible to observe relations between lines, angles and figures in the diagram, which are suggestive of the steps by which the necessary construction might have been effected.

This is called the Method of Analysis (áváλvois-resolution). It is a method of discovering truth by reasoning concerning things unknown or propositions merely supposed, as if the one were given or the other were really true. The process can best be explained by the following examples.

Our first example of the Analytical process shall be the 31st Proposition of Euclid's First Book.

Ex. 1. To draw a straight line through a given point parallel to a given straight line.

Let A be the given point, and BC be the given straight line. Suppose the problem to be effected, and EF to be the straight line required.

Now we know that any straight line AD drawn from A to meet BC makes equal angles with EF and BC. (1. 29.)

This is a fact from which we can work backward, and arrive at the steps necessary for the solution of the problem; thus:

Take any point D in BC, join AD, make ▲ EAD= L ADC, and produce EA to F: then EF must be parallel to BC.

Ex. 2. To inscribe in a triangle a rhombus, having one of its angles coincident with an angle of the triangle.

Let ABC be the given triangle.

Suppose the problem to be effected, and DBFE to be the rhombus.

Then if EB be joined, ▲ DBE= ▲ FBE.

This is a fact from which we can work backward, and deduce the necessary construction; thus :

Bisect ABC by the straight line BE, meeting AC in E. Draw ED and EF parallel to BC and AB respectively. Then DBFE is the rhombus required. (See Ex. 4, p. 59.)

Ex. 3. To determine the point in a given straight line, at which straight lines, drawn from two given points, on the same side of the given line, make equal angles with it.

Let CD be the given line, and A and B the given points. Suppose the problem to be effected, and P to be the point required.

B

D

We then reason thus:

If BP were produced to some point A',

▲ CPA', being= ▲ BPD, will be= ▲ APC.

Again, if PA' be made equal to PA,

AA' will be bisected by CP at right angles.

This is a fact from which we can work backward, and find the steps necessary for the solution of the problem; thus:

From A draw AO 1 to CD.

Produce AO to A', making OA'=0A.

Join BA', cutting CD in P.

Then P is the point required.

NOTE 10. On Symmetry.

The problem, which we have just been considering, suggests the following remarks:

If two points, A and A', be so situated with respect to a straight line CD, that CD bisects at right angles the straight line joining A and A', then A and A' are said to be symmetrical with regard to CD.

The importance of symmetrical relations, as suggestive of methods for the solution of problems, cannot be fully shewn

to a learner, who is unacquainted with the properties of the circle. The following example, however, will illustrate this part of the subject sufficiently for our purpose at present.

Find a point in a given straight line, such that the sum of its distances from two fixed points on the same side of the line is a minimum, that is, less than the sum of the distances of any other point in the line from the fixed points.

Taking the diagram of the last example, suppose CD to be the given line, and A, B the given points.

Now if A and A' be symmetrical with respect to CD, we know that every point in CD is equally distant from A and A'. (See Note 8, p. 103.)

Hence the sum of the distances of any point in CD from A and B is equal to the sum of the distances of that point from A' and B.

But the sum of the distances of a point in CD from A' and B is the least possible when it lies in the straight line joining A' and B.

Hence the point P, determined as in the last example, is the point required.

NOTE. Propositions IX., X., XI., XII. of Book I. give good examples of symmetrical constructions.

NOTE 11. Euclid's Proof of I. 5.

The angles at the base of an isosceles triangle are equal to one another; and if the equal sides be produced, the angles upon the other side of the base shall be equal.

Let ABC be an isosceles ▲, having AB= 4C

Produce AB, AC to D and E.

·· FA=GA, and AC=AB, and ▲ FAC= ▲ GAB,

.. FC=GB, and ▲ AFC= ▲ AGB, and ▲ ACF = ▲ ABG.

L

::: BF=CG, and FC=GB, and ▲ BFC= ↳ CGB,
.. 4 FBC= 4 GCB, and ▲ BCF= 4 CBG,

Now it has been proved that ACF= 2 ABG,
of which the parts BCF and 4 CBG are equal;

.. remaining ACB=remaining 4 ABC.
Also it has been proved that FBC= 4 GCB,
that is,
L DBC= L ECB.

I. 4.

Ax. 3.

Q. E. D.