SECTION II.

FUNDAMENTAL GEOMETRICAL PROPOSITIONS.

THEOR. 9. If there are two pairs of straight lines all of which are parallel, then the intercepts made by each pair on a straight line that cuts them are one to another in the same ratio as the corresponding intercepts on any other straight line that cuts them.

Let AB, CD and EF, GH be two pairs of straight lines all of which are parallel, and let AG, BH be two straight lines that cut them:

Suppose 2AC=3EG.

On AG take AA' equal to AC, so that CA'=2AC, and GG', G'G" each equal to EG, so that EG"=3EG.

Draw A'B', G'H', G"H" parallel to AB to meet BH in B', H', H”

If m.AC=n.EG, we can prove by like reasoning that m.BD

=n.FH.

Hence AC: EG:: BD : FH.

Def. 5. Q.E.D.

COR. I. If there are three parallel straight lines, the intercepts made by them on any straight line that cuts them are to one another in the same ratio as the corresponding intercepts on any other straight line that cuts them.

COR. 2. If the sides of a triangle are 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.

THEOR. IO. 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 in each case there is only one such point of division.

then can AB be divided internally and externally in the ratio H: K; and in each case in one point only.

From A draw the straight line AF making an angle with AB. From AF cut off AC equal to H, and CD, CE each equal to K; join BD, BE, and from C draw CP parallel to BD, and CQ to BE, meeting AB internally and externally at P and Q.

Because PC is parallel to BD,

therefore AP: PB:: AC: CD

therefore AP: PB::H: K.

Again, because QC is parallel to BE,

therefore AQ: QB:: AC: CE,

therefore AQ: QB:: H: K.

IV. 9, Cor. 2.

Hence AB is divided internally at P and externally at Q in the given ratio H: K.

Also AB cannot be divided internally in the ratio H: K at any point other than P.

For, if possible, let P' be a point such that AP': P'B:: H: K.

Join CP', and from B draw BD' parallel to CP' and meeting AF at D'.

Then, because CP' is parallel to D'B,

therefore AC: CD':: AP': P'B

but AP': P'B::H:K:: AC: CD

IV. 9, Cor. 2.

therefore AC: CD':: AC: CD

therefore CD'=CD

IV. 2.

which is impossible.

Hence P is the only point which divides AB internally in the given ratio.

A like demonstration will shew that Q is the only point which divides AB externally in the given ratio.

Q.E.D.

*Ex. 1. Trace the changes in position of P and Q, as K increases from being much smaller than H, till it is equal to H, and then till it is much larger than H.

DEF. IO. When a line AB is divided internally at P and externally at Q in the same ratio, it is said to be divided har monically by the points P and Q; and the points A, P, B, Q are said to form an harmonic range.

* Ex. 2. If A, P, B, Q form an harmonic range, and M is the middle point of AB, prove that MA or MB is the mean proportional between MP and MQ.

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

Let the straight line DE divide the sides AB, AC of the triangle ABC, so that AD: DB:: AE: EC,

B

then shall DE be parallel to BC.

For the parallel to BC through D divides AC in the ratio

AD: DB

IV. 9, Cor. 2.

and therefore must pass through E, the only point in which AC

Let AC, DF be two rectangles of equal altitude on the bases AB, DE: