but when ABC <rt. , then thoses are internal.
As the obtuse ABC decreases, the base DE diminishes
and the sides BD, BE approach;

when ABC becomes a rt. L, BD and BE coincide;
and after ABC becomes < a rt., BD and BE change
sides.

In the general proposition, this isosc. A DBE, is what

the perpendicular is, when the given A is rt. angled. .. the sides of this A DBE, and the triangles under them, and the sides of the given ▲ ABC, possess many of the properties already proved in the case of a rt. dA; for instance, as in Cor. 1, Pr. 8, VI.;

1. AC AB = AB : AD, where AB is a mean proportional. 2. AC CB = CB : CE, where CB is a mean proportional. AB = BC : BE,

3.

AC

4. AD BD BD : CE, where BD is a mean proportional.

5.

2.

:

or since BD = BE; AD: BE = BE: CE, where BE is a mean propl.

.' AD : BD = AB : BC, .. the segments AD and EC are in the duplicate ratio of the sides AB and BC.

"Hence in a right-angled triangle the segments of the hypotenuse by the perpendicular, are in the duplicate ratio of the sides." LARDNER'S Euclid, pp. 187, 188.

USE & APP. 1. The first Corollary of Pr. 8, bk. VI. supplies the principle on which a very clear and brief demonstration may be given of Prop. 47, bk. I.

2. Also according to this Proposition, and by aid of a square, inaccessible distances, as DB, may be measured.

At D raise the perp. DA, and measure it;

And at A place a square, so that by looking along one of its sides Ab, the point B may be seen in the same st. line with Ab;

And along the other side Ac, the point C may be seen; and measure DC;

Then, . CD : DA = DA : DB; .. DB =

Ex. Suppose AD = 3, and CD = 2.25; then DB =

DA2

CD

9

2.25

3. In a circle any chord, as BA, is a mean proportional between the diameter BC, and that segment of the diameter BD, which is drawn from one extremity of the chord, B, and cut off by a perpendicular, AD, let fall from A the other extremity of the chord.

'.' As BAC and ADB have each a rt., and B

common;

..the ABAC is eq. ang. to A ADB,

and.. BC: BA = BA: BD;

i. e. the chord BA is a mean propl. to BC & BD. And. As BAC, BDA and ADC are similar, the segments of the hypotenuse are in the duplicate ratio of the sides.

PROP. 9.-PROB.

From a given st. line to cut off any part required, i. e., any measure or submultiple.

CON.-3, I. 31, I.

DEM.-2, VI. 18, V. Componendo. If Ms, taken separately, be propls,
they shall also be propls when taken jointly; i. e, if first: 2nd
4th; then 1st + 2nd 2nd 3rd 4th 4th.

:

3rd:

D, V-If 1st : 2nd 3rd 4th, if 1st a m or pt of the 2nd, the 3rd is the same m or pt of the 4th.

Def. 1, V.-A less M is a pt of a greater when the less measures the

31, I.

4 Sol.

D. 1 C. 3.

22, VI.

3 18, V.

4 C. D, V.

Let AB be a given st. line;

to cut off from it any pt.
required.

From. A draw AC making any
Z with AB;

in AC take any. D, and make

AC the same m of AD
that AB is of AE the pt
to be cut off.

join BC, and draw ED || BC;

then AE is the submultiple required;
ED || BC one of the sides of ▲ ABC;

.. CD: DA = BE: EA;

and compon., CA : AD = BA : AE;

but CA is a m of AD; .. BA the same m of AE;

5 Def. 1,V... AE the same pt of AB, that AD is of AC.

6 Recap.

. From a given st. line, &c.

Q. E. F.

SCH. Prop. 10, Bk. I. by which a rt. line may be bisected, and its bisections also bisected, is a particular case of this Problem.

USE & APP. A simple extension of the Problem enables us; 1st, to divide a given line, AL, into any number of equal parts.

2.

Join EL, and through D, C, B draw ||s to EL and cutting AL;

then AL is divided into four eq. pts. in b, c, d, L.

...Ab is the same pt. of AL that AB is of A E;
i. e. the 4th part;

and ⚫. Ab, bc

cd dL;

AL is divided into four eq. parts.

To divide a triangle, ABC, into any number of eq. parts, say four, by lines from a given point, P, in one of the sides, as BC.

C. 1 Sch. 1. 9, VI. 2 31, I.

3 Pst. 1, I. 4 Sol.

Divide BC into four eq. pts. in D, E, F, and join AP;
and through D, E, F draw DG, EH, FK each || AP;
then join AD, AE, AF, and PG, PH, PK;
and the As BPG, GPH, HPK, KPC, each

=

AABC;

i. e. by lines from P, ▲ ABC is divided into 4 eq. pts

D. 1 37,

I.

A DPG: =A GAD, both being on GD and between the same s GD, AP;

2 Add. Ax. 2, I. to each add ▲ BGD, .. ▲ BPG = ▲ ADB.

3 C 1.

5 D. 3

6 Sim.

7 Sub.

Next, . BD = DE = EF = FC and the altitude com. .. As ABD, ADE, AEF, AFC, are equal;

and each is ▲ ABC;

but ▲ BPG ▲ ADB, ... BPG :

= ▲ ABC.

So A GPA ▲ DPA, and ▲ HPA = A APE,
take ▲ HPA from ▲ GPA, and ▲ APE from ▲ DPA,

8 Ax. 3, I. D. 4... rem. A GPH rem. ▲ AED = ▲ ABC.

Given the nth part of a line AB, to find the (n + 1) th part.

C. 1 46, I. Pst. 1, I. On AB desc. a sq. ABEF, and join AF, EC cutting in G;

through G draw HGK || AE, cutting AB, EF, in H & K; AB

then AH, or EK =—; or (n+ 1) AH

n+1

=

AB.

As AHG, FKG, are eq. ang.; and As AGC & EGF; AH: FK, or BH, AG: GF AC: EF, or AB. and .. BH = n. AH; and AB = (n+1) AH;'

To divide a given st. line similarly, i. e., proportionally, to a given divided st. line; or to divide a given st. line into parts that shall have the same ratios to one another which the parts of the divided given st. line have.

Or, “To divide a given undivided line similarly to a given divided line.”
EUCLID.

CON. Pst. 1, I. 31, I. 1, I.