DEM. D, VL 1, II. 14, VI.

D. 1 H.

2 D. VI.

3 H.

4 D, 2, 3.

5 1, II.

6

7 14, VI.

8 Conc.

9 Rec.

ADBC is a qu. lat. inscribed
in a O, of which AB and CD
are diagonals;

.. AD. CB + DB. AC=
AB. CD.

but AC = CB, .. AD. CB
= AD, AC

.. AD. AC + BD. AC =

AB. CD.

D

B

But AD. AC, & BD. AC are the rectangles
contained by AC & AD + DB;

.. rect. AC. (AD + DB) =rect. AB C. D;
&. the sides of eq. rectangles are reciprocally propl;
.. AD + DB: DC = AB: AC.

Therefore, if a segment of a circle be bisected &c.

Q. E. D.

SCH. This and the following Subsidiary Propositions have been adopted, with some slight alterations from BELL'S Plane Geometry, p. 194 - 198.

PROP G.-THEOR.

If two points, E & F, be taken in the diameter AC of a circle, or of the diameter produced, CF, such that the rectangle, ED.DF, contained by the segments intercepted between them and D the centre of the circle be equal to the square of AD the semidiameter; and if from these points two st. lines, EB, FB, be inflected to any point whatever, B, in the circumferenece of the circle, the ratio of the lines inflected, FB:BE, will be the same with the ratio of the segments, FA: AE, intercepted between the two first mentioned points and the circumference of the circle.

C

DEM. 17, VI. 6, VI. 4, VI. 16, V. altern. 17, V, dividendo. 11, V.

| Pst. 1 & 2, I. | Join DB, CB, AB & prod. FB to G.

COR. 1.

FD: DB = DB: DE;

& ·.· in ▲s FDB, BDE the sides about the
common D are propl.;

..▲ FDB is eq. ang. with ▲ BDE,
DBF = DEB, & DBE = / DFB.
▲

.. FB : BD = BE: ED,

& alt. FB: BE = BD: ED, or AD: DE.
But FD: DB or DA = DA: DE;
.. div. FA: DA = AE: ED.
& alt. FA: AEDA: ED.
Now FB: BE = AD: DE,
.. FB : BE = FA: AE.

• If two points be taken in the diameter, &c.

=

Q. E. D.

FB: BE FA: AE, .. on joining AB, by 3,VI., ZFBE is bisected by AB.

COR. 2. Also on joining BC, the ext. vert. EBG is bisected by BC; for

D. 117,VI.&18,V.| ·.· FD : DB or DC = DC: DE,

2 17, V.

3 G, VI.

4 A, VI.

.. comp. FC: DC = CE : ED;
& ·· FA : AD or DC = AE : ED,
.. ex. æq. FA: AE = FC: CE.
But FB: BE = FA: AE,
.. FB: BE =
FC: CE;
..ext, EBG is bisected by BC.

Q. E. D.

PROP. H.-THEOR.

If from one extremity, A, of AC the diameter of a circle ABC, a chord AB be drawn, and a perpendicular DE, to the diameter, cut both the diameter and the chord either internally or externally in D and F, the rectangle, CA. AD, under the diameter and its segment reckoned from that extremity A, is equal to the rectangle BA. AF, under the chord and its corresponding segment.

D. 1 H. 31, III.|

2 H. 15, I.

3

4 4, VI.

C D

= <DAF;

ZABC is in a semic, .. ABC is a rt. ; but ADF is also a rt. ▲, & ▲ BAC .. AABC is eq. ang. with ▲ ADF; ..BA: AC = AD: AF:

If the angles, A & B, at the base AB of a triangle ABC, be bisected by two lines, AD, BD, that meet, as in D, and the exterior angles at the base, formed by producing the two sides CA,CB, be similarly bisected by AG & BG; then the two points of concourse, D & G, and the vertex, C, shall be in one st. line, which shall bisect the vertical angle, ACB.

PRELIMINARY THEOREM, that may be demonstrated by superposition, “If two As have two sides of the one respectively equal to two sides of the other, and the opp. one of the sides in the first equal to the opp. to the equal side in the second, these As are equal when they are of the same species or affection, i. e., when they are both acute-angled, both right-angled, or both obtuseangled. See Sch. P, 7, VL

D. 1, Hyp.

2 26, I. 3 Sim.

4

5D, 2, 3.

6 D. 5, C.

7 Prel. Theor. 8 Sim.

9 Conc.

10 Rec.

Draw DE, DF, DL to the sides

AC, BC & AB;

& GM, GN

to the sides produced AM, BN;

& GK to the side AB.

A

M

B

N

in As ADE, ADL, EAD = DAL,
Zs at E & L also equal, & AD common.
AE & DL = DE.

.. AL
So BL = BF, & DL =

DF;

also AM = AK, GM = GK, BN =
& GN = GK.

& DE & DF each = DL,

BK,

.. DE = DF, & sim. GM = GN
Again, in As CED, CFD, CD,DE=CD,DF,
& Ls at E & F are rt. 8;

.. ▲ CED = ▲ CFD, & / ECD = / FCD.
So ACMG ▲ CNG, & ▲ MCG = NCG;
.. the lines CG & CD coincide.

. If the angles at the base, &c.

Q.E. D.

PROP. L.-Theor.

In a triangle ACB, as in the last proposition, the segments, CM or CN, of each side produced that are intercepted between the vertex, C and the external perpendiculars, GM, GN, are each equal to S, the semiperimeter of the triangle; the segments, CF or CE, of these sides next the vertex, C, are equal to S—AB, the excess of the semiperimeter above the base AB; and the segment AE or BF, of each of these sides next the base is respectively equal to S—BC, or to S—AC, the excess of the semiperimeter above the other side.

DEM. 26, I. Ax. 6 & 7, I. Ax. 2, I.

D. 1 26, 1.

2 26, I.

3 Ax. 6, I.

4D. 5, K. D. 2, K
5 Ax. 6 & 7, I.
6 Add. Ax. 2.
7 D. 5 & 4.
8 Conc. 1

9 D. 7, 6 & K, 4

10 Conc. 2.

11 D. 7,& K 3.D.2 12 Conc. 3.

13 Rec.

And ·.· AE =AL = BK =

.. AE = S BC;

AB,

BN;

&.. AM=AK=BL=BF, & AM=S-AC, .. BF = S-AC

i. e., each seg. of the side next the base is equal to the excess of the

above the other side.

the segment of each side, &c.

semiperimeter

Q. E. D.