PROPOSITION IO.

"If four magnitudes of the same kind be proportionals, the first is greater than, equal to, or less than, the third, according as the second is greater than, equal to, or less than, the fourth."

then A> or < C according as B> or < D.

If B > D, then

=

"If four magnitudes of the same kind be proportionals, the first will have to the third the same ratio as the second to the fourth." (Alternando.)

Let

A B C D, then will A: C: B : D.

:

::

D.-SEVEN FUNDAMENTAL THEOREMS IN PROPORTION.* PROPOSITION 12: (Lemma).**

If on two straight lines, AB, CD, cut by two parallel straight lines AC, BD, equimultiples of the intercepts respectively be taken; then the line joining the points of division will be parallel to AC, or BD.”

On A B and CD, produced either way, let the respective equimultiples B E, DF of A B, C D be taken, on the same side of BD; then EF is parallel to BD.

Since the

For, join A D, DE, BC, BF. triangles ABD, CBD are on the same base BD, and their vertices A, C are in the line AC parallel to BD, they are equal in area; and whatever multiple BE is of AB, or DF of CD, the triangle DBE is that same multiple of the triangle ABD, and the triangle DBF of the triangle CBD.

... area of triangle EBD = area of triangle FBD.

E

B

F

Fig. 3

But these triangles EBD, FBD have the same base BD; hence their vertices E and F must be in a straight line parallel to BD and therefore

EF is parallel to BD.

* Enunciations of Propositions 13-17 quoted from the Syllabus of Plane Geometry, Book IV, Section 2.

** J. M. Wilson: Elementary Geometry, page 205.

PROPOSITION 13.

If two straight lines be cut by three parallel straight lines, the intercepts on the one are to one another in the same ratio as the corresponding intercepts on the other."

Let the three parallel straight lines A A', B B', CC' be cut by two other straight lines AC, A'C' in the points A, B, C and A', B', C' respectively; then

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

For, on AC take B M = m. AB, BN: =n. BC, m and

n being integers, M and N on the same side of B.
A'C' take B' M' =m. A'B', B'N' =n. B'C',
M' and N' being on the same side of B' as M
and N of B. Then, by the foregoing lemma
(Prop. 12), M M' and N N' are both parallel to
B B' and cannot meet. Hence, whatever in-
tegers m and n may represent,

B' M' (or m. A' B')> or < B'N' (or n. B'C')
according as

=

B M (or m, AB)> or < BN (or n. BC); ... AB: BC:: A'B': B' C'.

N

B

Also on

A

Fig. 4

M'

(i). COROLLARY: "If the sides of a triangle be cut by a straight line parallel to the base, the segments of one side are to one another in the same ratio as the segments of the other side."

(ii). COROLLARY: "If two straight lines be cut by four or more parallel straight lines, the intercepts on the one are to one another in the same ratio as the corresponding intercepts on the other."

(iii). COROLLARY: If in any triangle, as OAB, a straight line EF, parallel to the side A B, cut the other sides, OA in E and OB in F, then

AB: EF:: OA: OE :: OB: OF.

PROPOSITION 14.

“A given finite straight line can be divided internally into segments having any given ratio, and also externally into segments having any given ratio except the ratio of equality;" and if the line be given in both length and sense, there is in each case one and only one such point of division.

Let A B be the given straight line; it may be divided, as at E, in a given ratio P : Q.

For, on the straight line AG making any convenient angle with AB lay off ACP, CD=Q. drawn parallel to DB to meet

AB in E, will divide A B at E in the given ratio. (Prop. 13.)

Since CE and D B are parallel, C and E lie on the same side of D and B, and hence the A division will be internal if A and D are on opposite sides of C, but external if A and D are on the same side of C.

FE B

G

Fig. 5

Then CE,

G

B

F E

If the line to be divided be estimated in a given sense, as from A to B, there is in each case only one point of division in the given ratio. For if any other point, as F, be joined to C and BG be drawn parallel to FC, then AF: FB: AC: CG,

(Prop. 13.) so that F divides A B in the ratio AC: CG, different from the given ratio.

If the given ratio be a ratio of equality, the construction in the case of external division fails.

PROPOSITION 15.

"A straight line which divides the sides of a triangle proportionally is parallel to the base of the triangle."

Let DE divide the sides A B, AC of the triangle ABC proportionally, so that

then

AD: DB:: AE: EC,

DE is parallel to BC.

If possible, let DF be parallel to BC, E some other point than E; then

AD: DB:: AF: FC.

But by hypothesis

(Prop. 13.)

AD:DB:: AE: EC

... AF: FC:: AE: EC

which is only possible when F coincides

with E.

FROPOSITION 16.

'Rectangles of equal altitude are to one another in the same ratio as their bases.'

Let KA, KB be two rectangles having the common altitude OK and their bases OA, OB extending in the same line from O to the right; then

rect. KA rect. KB :: OA: OB.

KN. Whatever multiples OM and ON are of OA and OB, the rectangles K M and KN are the same respective multiples of the rectangles K A and KB; that is,