BOOK VI.

RATIO APPLIED.

I. Fundamental Geometric Proportions.

THEOREM I.

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

HYPOTHESIS. Let the three parallols AA', BB', CC', cut two other lines in A, B, C, and A', B', C', respectively.

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

PROOF. On the line ABC, by laying off m sects = AB, take BM = mAB, and, in the same way, BN = n.BC, taking M and N on the same side of B. From M and N draw lines || AA', cutting A'B'C'

in M' and N'.

:. B'M' = m. A'B', and

B'N' = n. B'C'.

(227. If three or more parallels intercept equal sects on one transversal, they intercept equal sects on every transversal.)

But whatever be the numbers m and n, as BM (or m. AB) is >, =, or < BN (or n. BC),

so is B'M' (or m. A'B') respectively >, =, or < B'N' (or n. B'C'); :. AB : BC :: A'B' : B'C'.

498. REMARK. Observe that the reasoning holds good, whether B is between A and C, or beyond A, or beyond C.

B

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B'

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499. COROLLARY I. If the points A and A' coincide, the figure ACC' will be a triangle; therefore a line parallel to one side of a triangle divides the other two sides proportionally.

500. COROLLARY II. If two lines are cut by four parallel lines, the intercepts on the one are to one another in the same ratio as the corresponding intercepts on the other.

501. If a sect AB is produced, and the line cut at a point P outside the sect AB, the sect AB is said to be divided externally at P, and AP and BP are called External Segments of AB.

In distinction, if the point P is on the sect AB, it is said to be divided internally.

THEOREM II.

502. A given sect can be divided internally into two seg ments having the same ratio as any two given sects, and also externally unless the ratio be one of equality; and, in each case, there is only one such point of division.

A FG

B

GIVEN, the sect AB.

On a line from A making any angle with AB, take AC and CD equal to the two given sects. Join BD.

Draw CF || DB and meeting AB in F. By 497, AB is divided internally at F in the given ratio.

If it could be divided internally at G in the same ratio, BH being drawn | CG to meet AD in H,

AG would be to GB as AC to CH, and therefore not as AC to CD. (476. The scale of relation of two magnitudes will be changed if one is altered in size ever so little.)

Hence F is the only point which divides AB internally in the given

ratio.

Again, if CD be taken so that A and D are on the same side of C, the like construction will determine the external point of division. In this case the construction will fail if CD = coincide with A.

AC, for D would

As above, we may also prove that there can be only one point of external division in the given ratio.

503. INVERSE OF 499. A line which divides two sides of a triangle proportionally is parallel to the third.

For a parallel from one of the points would divide the second side in the same ratio, but there is only one point of division of a given sect in a given ratio.

THEOREM III.

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

Let AC, BC, be two rectangles having the common side OC, and their bases OA, OB, on the same side of OC.

=

In the line OAB take OM m.OA, and ON = n. OB, and complete the rectangles MC and NC.

Then MC = m. AC, and NC = n. BC; and as OM is >, =, or < ON, so is MC respectively >,=, or < NC;

.. rectangle AC rectangle BC :: base OA: base OB.

505. COROLLARY. Parallelograms or triangles of equal altitude are to one another as their bases.

THEOREM IV.

506. In the same circle, or in equal circles, angles at the center and sectors are to one another as the arcs on which they stand.

Let O and C be the centers of two equal circles; AB, KL, any two arcs in them.

Take an arc AM = m. AB;

then the angle or the sector between OA and OM equals m.AOB. (365. In equal circles, equal arcs subtend equal angles at the center.)

Also take an arc KN = n. KL; then the angle or sector between CK and CN equals n. KCL.

But as AOM>, =, or < KCN, so respectively is arc AM >, =, or < arc KN;

(370 and 372. In equal circles, equal angles at the center or equal sectors intercept equal arcs, and of two unequal angles or sectors the greater has the greater arc.)

:. AOB: KCL :: arc AB : arc KL.