But, thirdly, we can draw the line AM bisecting the angle D A E.

And this clearly gives us the solution of our problem, since

angles to A M.

we

can can now draw DPE at right Thus the solution runs as follows:

Draw A M bisecting the angle D A E, and through P draw DP ME at right angles to AM; then

For, in the triangles MAD is equal to the

shall AD be equal to A E. MAD, MAE, the angle angle MA E, the right angle AM D is equal to the right angle A ME, and AM is common to the two triangles; therefore the triangles are equal in all respects (Euc. I., 26), and A D is equal to A E.

The proof of the equality of the triangles M AD and M A E was not included in the prior examination of the problem, since it is involved in the assumed knowledge on the student's part of the fundamental properties of isosceles triangles, proved farther on. But of course it is well (in a case of such simplicity) to introduce the proof into the solution of the problem.

Let us next try the following problem :—

Ex. 8.-The points P and Q (Fig. 13) are on the same side of the line AB. It is required to determine a point C in AB, such that the lines PC, QC may make equal angles with A B.

Let C be the required point, so that the angle PCA is equal to the angle QC B.'

1 Construct as follows: Draw A B, and from any point C in A B draw the unequal lines C P, C Q equally inclined to A B.

Let us try drawing a line, C D, at right angles to A B. Then the angle PCD is equal to the angle QCD. On a consideration of this relation, however, it seems unlikely to help us. For it is not easier to gather anything from the equality of PCD and QCD, than to make use of the equality of PCA and QC B.

It seems an obvious resource, since the equality of the angles PCA and QC B, as they stand, is not

E

F

FIG. 13.

readily applicable to our purposes, to produce either PC or QC, in order to see whether the vertical angle either of PCA or QCB might be more serviceable to us. Produce PC to E. Then the angles QCB and BCE are equal, or CB is the bisector of the angle QC E. The only property connected with the bisector of an angle which seems likely to help us is this one, that the bisector of the vertical

Then there is no risk that accidental relations will appear as necessary ones.

angle of an isosceles triangle is perpendicular to and bisects the base. Now, we can make an isosceles triangle of which C shall be the vertex and CQ a side, for we have only to take CE equal to C Q, and to join Q E, cutting C B in F. Then, by the property just mentioned, QE is at right bisected in F.

angles to C F, and is

These relations obviously supply all we want. For, reversing our processes, we have only to draw QFE perpendicular to A B, and to take F E equal to QF; then drawing PE to cut A B in C, we are certain that C is the required point. In all such cases we should not be equally certain that the proof would be as simple as the analysis, since sometimes the reversal of a process involves properties not so readily seen as their converse theorems. In this case, however, it is obvious (or will at least appear so on a moment's inquiry) that the proof is simple.

For, join CQ (we are going now through the synthetic treatment of the problem, and therefore ignore the prior constructions), then, because QF is equal to FE, and CF is common and at right angles to QE, the triangles CFQ and CFE are equal in all respects. Therefore, the angle QCF is equal to the angle ECF. But ECF is equal to the vertical angle PCA. Therefore the angle QCF is equal to the angle PCA.

It is an excellent practice, when a problem has been solved, to notice results which flow from, or are in any way connected with, our treatment of the

problem. In Ex. 8 we notice that the line CQ (Fig. 13) is equal to the line CE, so that the sum of the lines PC, CQ is equal to the line PE. It might occur to us to inquire what is the sum of lines drawn from P and Q to any other point, as G, in A B. Join PG and Q G. The fact that CE is equal to CQ reminds us that if we join G E, GE will be equal to GQ. Thus PG and GQ are together equal to PG and GE together. But PG and G E are together greater than PE; that is, P G and G Q are together greater than PC and CQ together; or PCQ is the shortest path from P to Q, subject to the condition that a point of the path shall lie on A B.

VI. PROBLEMS ON MAXIMA AND MINIMA.

The result last obtained fitly introduces us to an important class of problems-viz. those in which we have to show that certain lines, areas, &c., are the greatest or least which can be constructed under certain assigned conditions. There are few problems of this sort in Euclid. In fact, the seventh and eighth propositions of the third book are the only theorems in Euclid expressly dealing with geometrical maxima and minima. But many interesting deductions involve such relations as we are speaking of, and it is well for the student to know how to deal with them.

It will be noticed that some of the problems already dealt with may be presented as examples of

geometrical maxima and minima. For instance, Ex. 4 may be presented in the following form :

Ex. 9.-From a point within a quadrilateral, lines are drawn to the angles of the quadrilateral: show that the sum of these lines will be a minimum when the point is at the intersection of the diagonals.

Presented in this form the problem would be solved precisely as Ex. 4. But suppose it had been given in the following form:

Determine a point within a quadrilateral such that the sum of the lines from the point to the angles of the quadrilateral shall be a minimum.

A

B

FIG. 14.

Here, assuming the student to have no knowledge of the property established in Ex. 4, the problem is not quite so simple. Let us see how it is to be

dealt with.

Draw first the quadrilateral ABCD (Fig. 14), and from some assumed point, P, draw PA, PB, PC and PD. Then we have to inquire how to shift