each other as two whole numbers, as 7 to 11 for instance, so that

AE: EB::7:11,

then will DF and FC be to each other in the same ratio.

For, if we conceive the straight line AB divided into 7+11=18 equal parts, and through the points of division parallels to BC to be drawn, they will divide DC into 18 equal parts (T. I.), of which 7 will belong to DF, and 11 to FC. Hence,

DF: FC::7:11.

Secondly. When AE and EB are incommensurable.

We may conceive AB divided into equal parts so small as to bring K, one of the points of division, as near as we please to E. Then drawing KL, we have, by the first case,

AK: KB::DL: LC.

And as this will continue, however small the divisions are, wo shall have (B. II., P. XIII., S.)

AE: EB:: DF: FC.

THEOREM III.

If a line be drawn parallel to either side of a triangle, cutting the other sides, it will divide them into proportional parts.

Although this Theorem might have been made a corollary to the last, we have preferred making it one of our principal Theorems, not only on account of its own importance, but also by reason of the important corollaries which it is capable of furnishing.

Let DE be drawn parallel to the side BC of the triangle ABC, cutting the other sides at D and E; then we shall have

AD: DB:: AE: EC.

Through A draw AG parallel to BC, and through any point, as G, of this parallel, draw

A G

I

GK parallel to AC meeting BC produced. Also produce DE to F. We have (T. II.)

AD:DB::GF: FK;

but, by reason of the parallels AC, GK, we have GF=AE, FK-EC; consequently,

AD:DB::AE: EC.

Cor. I. Conversely. If a straight line DE divide two sides of a triangle into proportional parts, it will be parallel to the third side. For, if this is not the case, we might draw through D another straight line DI parallel to BC, and we should then have AD: DB:: AI: IC;

but by hypothesis,

AD:DB:: AE: EC;

hence, by equality of ratios, we have

AI:IC:: AE: EC,

which is absurd, since AI<AE, and IC>EC.

Cor. II. From the proportion

AD:DB::AE:EC

we have, by composition,

AD+DB: AD::AE+EC: AE,
AD+DB: DB:: AE+EC: EC,

which gives these new propositions:

AB: AD:: AC: AE and AB:DB::AC: EC.

Conversely. If a straight line DE is so drawn as to give

AB: AD:: AC: AE,

we shall have, by division,

that is,

AB-AD: AD::AC-AE:AE,

DB: AD:: EC: AE,

which is the same as AD: DB::AE:EC;

consequently, the straight line DE is parallel to BC (C. I.).

Scholium. Since, in every proportion, the product of the extremes is equal to the product of the means, we derive from the above proportion, AD: DB:: AE: EC, this equation, ADXEC=DB×AE.

Now, in order to comprehend the sense which we ought to attribute to the expression, the product of two lines, AD× EC, or DBX AE, it is necessary to suppose that the straight lines AD, EC, DB, and AE, have been referred to the same linear unit; and instead of lines, properly speaking, we have only to consider, in reality, abstract numbers expressing the ratio of these lines to their unit. We ought then, as in Arithmetic, to attach to the word product the idea of the multiplication of two numbers.

Hereafter, we may be required to multiply a surface by a line, a surface by a surface, and even a volume by a volume, etc. But in all these cases it must be understood that it is the ratios of these geometrical magnitudes to their respective units, which are thus multiplied.

SIMILAR TRIANGLES.

THEOREM IV.

A straight line drawn parallel to either side of a triangle, intersecting the other two sides, will form a new triangle similar to the first.

If DE is parallel to BC, we shall have
AB:AD::AC:AE::BC: DE.

For, since DE is parallel to BC, we have

(T. III., C. II.)

AB: AD::AC: AE.

Drawing EF parallel to AB, we also have

AC: AE:: BC: BF or DE;

hence, we have, by equality of ratios,

A

D

E

B

AB: AD::AC: AE::BC: DE.

Consequently, these triangles are similar (D. II.).

THEOREM V.

Two similar triangles, that is to say, which have their sides proportional, have their homologous angles equal, each to each.

Let ABC and A'B'C' be the two A similar triangles.

=

A'

B"

C"

B

C B'

C'

Take AB" A'B', and draw B"C" parallel to BC; the two triangles A'B'C', AB"C", being each similar to ABC, the one by hypothesis, and the other by the preceding Theorem, are similar to each other, and as we have, by construction, AB"= A'B', these triangles must be equal, since the ratio of similitude is unity (D. I.). Now, since B"C" is parallel to BC, the angles of the triangle AB"C" are respectively equal to the angles of the triangle ABC; that is to say, we have the angle at A common, B"=B, C" C. Then, also, we must have A'= A, B'=B, C'=C.

We see that the equal angles are opposite the homologous sides.

THEOREM VI.

Two triangles, which have the angles of the one respectively equal to the angles of the other, are similar, and the sides opposite the equal angles are homologous.

In the two triangles ABC, A'B'C', suppose A=A', B=B', C= C'.

B

A

C"

A'

As in the preceding Theorem, take AB" A'B' and draw B"C" parallel в" to BC. The two triangles A'B'C, AB"C", are equal, having a side equal, A'B'=AB", and the adjacent angles equal, A'=A, B'=B=B". Now the triangle AB"C" is similar to the triangle ABC (T. IV.); hence its equal A'B'C' will be similar to ABC. Consequently we have this series of equal ratios:

AB: A'B':: AC: A'C':: BC: B'C'.

C

B'

C'

The homologous sides, AB and A'B', AC and A'C', BC and B'C', as we see, are opposite the equal angles C' C, B' B, A' A.

THEOREM VII.

Two triangles are similar when the sides of the one are respectively parallel or perpendicular to the sides of the other, and the homologous sides are the parallel or the perpendicular sides.

From the preceding Theorem we see that the demonstration will be made out if we can show that these triangles have their angles respectively equal.

Let AB and A'B', AC and A'C', BC and B'C', represent the respective sides, either parallel or perpendicular, we are to show that C=C', B=B', A=A'.

In effect, we know (B. I., T. III., C.) that the angles C and C', B and B', A and A', must be either equal or supplementary. Now, in the first place, it is impossible that three set of angles should be supplementary, since we should then have

A+A'+B+B'+C+C' 6 right angles,

which is absurd (B. I., T. V., C. I.).

Neither can two set of these angles be supplementary; as, for example, A and A', B and B', for then we should have

A+A'+B+B'= 4 right angles,

which is also absurd.

It is then necessary that two angles, at least, of the first triangle should be equal respectively to two angles of the second; ; consequently, the third angle of the first triangle must equal the third angle of the second (B. I., T. V., C. III.). So that we shall have

A = A', B = B', C = C' ;

and consequently the triangles are similar, and we have this series of equal ratios (T. VI.):

BC: B'C':: AC: A'C'::AB: A'B'.

From the method of demonstration we see that the homologous sides BC and B'C', AC and A'C', AB and A'B', are the parallel or the perpendicular sides.