XI. PROBLEMS ABOUT SHAPE.

At present there remains only one class of deductions to deal with-viz. those in which questions of shape are involved. There are many problems which, although sufficiently simple and easy, do not admit of being solved without a reference to the sixth book of Euclid; and there are others which are much more readily solved by means of the sixth book than without its aid.

Consider, for instance, the following example:

Ex. 24.-ABC (Fig. 37) is a given angle; it is required to draw a line, B D, so that when through any point D in BD a line, A DC, is drawn at right angles to BD, DC shall be equal to three times A D.

We

Here is a problem clearly depending on shapefor instance, not on the length of BD or AC. see that if we can divide any angle equal to ABC in the required manner our problem is solved; or, rather, we have to construct a figure resembling ABCD as A B CD is supposed to be drawn.

We notice that A D is one-fourth of A C, and D B at right angles to A C. We therefore draw a straight line E F(Fig. 38), take HF equal to one-fourth of E F, and draw H G at right angles to EF. All that is now required is that we should determine G so that the angle E G F may be equal to the angle ABC. This is readily effected, since we know how to describe on EF an arc EGF containing an angle equal to the angle ABC (Euc. III., 33); the intersection of the line HG with the arc EGF gives us the required

FIG. 38.

H

point G. We join E G, G F; then the angle EGF is equal to the angle A B C.

Now if we draw B D so that the angle A B D is equal to the angle H G F, then the remaining angle DBC is equal to the remaining angle HG E. Through any point D in the line B D thus obtained draw A DC at right angles to BD. Then obviously the triangle A B D is similar to the triangle G HF; therefore A D is to BD as FH to HG. Similarly

BD is to D C as G H to HE; therefore, ex æquali, A D is to D C as HF to H E. But HF is one-fourth of EF; therefore HE is equal to three times H F; and hence DC is equal to three times A D.-Q.E.F.

We do not give the considerations which lead to the last lines of the proof. The considerations respecting shape which led to the construction may be looked on as obvious, although (as is often the case) it may not be quite so obvious how the proof is to be made to depend on properties established in the sixth book.

In a problem of the above type we cannot well avoid the use of the sixth book. I now give a problem which can be solved by the third book, but one can scarcely doubt that the solution depending on the sixth book is that which would naturally occur to a person dealing with the problem as a new one :

Ex. 35.-Let AB, A C (Fig. 39) be two straight lines meeting in A, D a given point. It is required to draw a circle which shall pass through the point D and touch the lines A B, A C.

We first notice that the circle must have its centre in the line A E, which bisects the angle BA C. For, taking any point, F, on this bisector, and drawing perpendiculars FH and FG on A B and A C respectively, we see that FH is equal to FG (since the triangles FAH, FAG are equal in all respects, Euc. I., 26). We describe a circle H G K, with centre F and distance FH or F G, and touching A B and AC in H and G (Euc. III., 16). But this circle

does not pass through D. It is obvious, however, that if we draw A K D, cutting the circle H G K in K, then the figure formed by the lines A H, A G, the point K, and the circle H G K exactly resembles that which will be formed when our problem is solved. If, then, we can only form a figure resembling that we have constructed, but such that the circle shall pass through D, the problem will be solved. Now we see that F lies in a defined direction with respect

M

N

R

FIG. 39.

to K-in other words, the angle A KF does not vary with the size of the circle drawn as HGK was drawn. We have then only to draw DL so that the angle ADL is equal to the angle A KF-that is, we have only to draw DL parallel to K F, to determine L, the centre of the circle we require. The proof runs thus :

Draw LM and LN perpendicular to AB and AC respectively. Then, from the similar triangles

E

ALD, AFK, LD is to A L as FK to AF. Again, from the similar triangles, A LN, AFG, AL is to LN as AF to F G. Therefore, ex æquali, LD is to LN as FK to FG. But FK is equal to F G; therefore LD is equal to LN-that is, to LM. Hence a circle described with centre L at distance LD will pass through M and N, and touch A B and AC in these points respectively, since the angles at M and N are right angles.

NOTE. Of course there is no difficulty in solving this problem without the aid of any book beyond the third. The obvious course of proceeding is by way of analysis. Let D N M be the required circle, having its centre at L. ALE is the bisector of the angle BAC and can be drawn at once. RDO PQ at right angles to AE can also be drawn, and gives P, a point on the required circle; for DO-OP. Then one can hardly miss the relation that the square on RN is equal to the rectangle under RD, R P; whence N is given, and the required circle, passing through the three known points, P, D, N, is determined.--Q.E.F.