Sidebilder
PDF
ePub
[merged small][ocr errors][merged small][merged small][merged small][merged small]

THEOREM .

EQUAL parallelograms (A B, BC), which have one angle of the one

[ocr errors]

(F B D) equal to one angle of the other (GBE), have their fides (FB, BD & GB, B E), about the equal angles reciprocally proportional, (that is, FB: BE = GB: BD). And parallelograms that have one angle of the one (FBD) equal to one angle of the other (G B E) and the sides (F B, BD & G B, B E), about the equal angles reciprocally proportional, are equal. Hypothesis.

Thesis. 1. The pgr. A B is = 10 the pgr. B C.

FB : BE = GB : BD. II. V F B D is = to Y G B E.

Preparation.
1. Place the two pgrs. A B, B C fo as the fides FB, B E

may be in a straight line F E.
2. Complete the pgr. D E.

Pof.2.B.1. I DEMONSTRATION. ECAUSE the y FB D, G Be are equal (Hyp. 2); & FB,

B E are in a straight line FE (Prep. 1). 1. Therefore, GB, B D are in a ftraight line G D.

P.14. B. 1. But the pgr. A B being = to the pgr. B C (Hyp. 1). 2. The

Pgr.
AB:
Pgr.
DE
Pgr.
DE

P. 7. B. 5. But the pgrs. AB, DE also BC, DE have the same altitude (D.4.B.6). 3: 'Hence

Pgr.

DE=FB:BE & k pgr. BC: pgr. DE=GB: BD. P. 1. B.6. 4. Consequently, FB:BE = GB : B D ( Arg. 2).

P.11.B.5.
Which was to be demonstrated.
Hypothesis.

Thesis. 1. FB; BE=GB: B.D.

The pgr. A B is = to the pgr. B C. II. V FB Dis = to GBĖ.

II. DEMONSTRATION. T

may be demonstrated as before, that GB, BD are in the line GD. But the pgrs. AB, DE, & BC, DE, have the same altitude (D.4.B.6). 2. Hence, pgr. AB : pgr.DE=FB : BE, & pgr. BC: pgr.DE=GB: BD. P. 1. B.6.

But FB:BE=GB: BD (Hyp.).
3. Wherefore, the pgr. A B : pgr. D E = pgr. B C : pgr. D E.
4. Consequently, the pgr. A B is = to the Pgr.

BC.
Which was to be demonstrated.

B C : pgr.

Pgr. AB :

1.IT

P.U. B.5. P. 9. B.5.

[merged small][merged small][ocr errors][merged small][merged small]

PROPOSITION XV. THEOREM X. EQUAL triangles (AC B, E C D) which have one angle of the one

(m) equal to one angle of the other (n): have their sides (A C, C B, & EC, CD), about the equal angles, reciprocally proportional ; & the triangles (ACB, ECD) which have one angle in the one (m) equal to one angle in the other (n), and their sides (A C, CB, & EC, CD), about the equal angles reciprocally proportional, are equal to one another.

CASE I.
Hypothesis.

Thelis. 1. The A ACB is = to AECD.

The fides AC, CB & EC, CD, II. Vm is = to y n.

are reciprocally proportional, or

AC: CD = EC: CB.

Preparation.
1. Place the A ACB, ECD so that the Gides A C, CD

may be in the same straight line A D.
2. Draw the straight line BD.

Pof.1. B.1.

DEMONSTRATION.

BECAUS

1.

P.14. B.I.

2.

ECAUSE Vm= yn (Hyp. 2.), & the straight lines AC, CD
are in the same straight line AD (Prep. I).
The lines E C, C B are also in a straight line E B.
But the A AC B being = to the AECD (Hyp. 1).
The A ACB:ACBD=AECD:ACBD.

P. 7. B.5. But the AACB, CB D also E CD, CB D have the same altitude

(Prep. 2. Arg. 1. & D. 4. Rem. B. 6). 3. Wherefore the A ACB : ACBD = AC:CD.

P. 1. B.6. &

the AECD: A CBD = EC:CB. 4. Consequently, AC: CD= EC: CB (Arg.2.&P.11.B.5).

Which was to be demonstrated.

[blocks in formation]

CASE II.
Hypothesis.

Thesis. 1. AC: CD= EC:CB.

The AACB, is in the AECD. II. & Vm = V n.

Preparation.
1. Place the two A ACB, E C D so that the sides AC, CD,

may be in the fame straight line A D.
2. Draw the straight line BD.

DEMONSTRATION.

1.

2. The

T may be demonstrated, as in the first Case, that EC, CB are in
the same straight line E B.
And because the A ACB, CBD, also the AECD, CB D have
the same altitude (Prep. 2. Arg. 1. & D. 4. Rem. B. 6).
A ACB : ACBD=AC: CD.

P. 1. B.6.
Likewise A ECD: ACBD = EC: C B.
But

AC: CD=EC: CB. (Hyp.1). 3. Wherefore A ABC: ACBD = A ECD: ACBD 4. Consequently, the AABC is = to the AECD

P. 9. B. 5. Which was to be demonstrated.

P.U. B.5.

[ocr errors]
[ocr errors]

2. Therefore the sides of the rgles E B, F D about the equal VA&7. 01. B.5.

N

M
E

C
F.

H
A
B с

D
PROPOSITION XVI. THEOREM XI.
F four straight lines (A B, CD, M, N) be proportionals, the rectangle con-
tained by the extremes (A B. N) is equal to that of the means (C D. M). And
if the rectangle contained by the extreames (A B. N) be equal to the re&tan-
gle contained by the means (CD. M), the four straight lines (AB, CD, M, N)
are proportionals.
Hypothehs.

Thesis.
AB:CD= M: N.

Rgle. A B. N = Rgle. C D. M.

Preparation. 1. At the extremities A & C, of AB,CD, erea the I AE,CF. P.11. B.5. 2. Make A E=N, & CFM.

P. 3. B.I. 3. Complete the rgles. E B, F D.

P.31. B.1.
I. DeMONSTRATION.
ECAUSE AB:CD=M:N (Hyp.):& M=CF & N=AE (Prep.2).
AB : CD = CF; AE.

,
(Prep. 1. & Ax. 10. B. 1.) are .

D. 2. B.6. 3. Consequently, the rgle. EB = rgle. FD, or the rgle under AB.AE SP.14.B.6. = the rgle. under CD. CF.

{D. IB.1. Consequently, A E being =N&CF=M (Prep. 2). 4. The rgle. under A B. N is also = to the rgle. under CD. M. Ax.2. B.2.

Which

was to be demonstrated.
Hypothesis.

Thesis.
The rgle. AB. N is = to the rgle. CD.M.

AB:CD=M:N.
II. DEMONSTRATION.

rgle. N AE=N, & CF=M (Prep. 2). 1. The igle. under A B. A E is = to the rgle under CD.CF. Ax.2. B.s. But these fides being about the equal ÞEAB, FCD (Prep.1. & AX.10.B.1). AB; CD= CĖ: A E.

P.14. B.6.
And

CF being = M & A ESN (Prep. 2).
AB:CB = M: N.

P.7. && 11. B.5,
Which was to be demonstrated.
Ee

Because

P.7

[ocr errors]

3.

[blocks in formation]

(F three straight lines (AB, CD, M) be proportionals, the re&angle (AB.M) contained by the extremes is equal to the square of the mean (CD): And if the re&tangle contained by the extreams (AB.M) be equal to the square of the mean (CD), the three straight lines (AB, CD, M) are proportionals. Hypothefis.

Thesis. AB :CD=CD: M.

The rgle. AB.M is = to the of CD

Preparation 1. At the extremities B & D of AB, CD erect the IBE, DF. Pu. B.s. 2. Make B E = M & DF =DC.

P. 3. B.1. 3. Complete the rgles. E A, FC.

P.zi. B.1. 1. DEMONSTRATION. ECAUSE AB: CD=M (Hyp.), & CD=DF & M=BE (Prep. 2) AB:CD=DF:BE.

P.7. &U.B.S, herefore the sides of the rgles. E A, FC about the equal V B & D (Prep. 1. & Ax. 10. B. 1) are reciprocal.

D. 2. B.6 2. Consequently, the rgle. E A is = to the rgle. FC, or the rgle. under A B. BE= the rgle Ç D. DF.

SP.14. B.1. 3. Wherefore, B e being = M&DF=CD (Prep. 2), the rgle. D. 1.B.6. A B. M is also = to the O of C D.

Ax.
Hypothesis.

Thesis.
The rgle. A B. M is = to the o of CD.

AB:CD=CD:M II. DeMONSTRATION. BECAUS:

ECAUSE the rgle. under AB.M is = to the of CD (Hyp.), & that B E is =M&DF=CD (Prep. 2). 1. The rgle. under A B. B E is = to the rgle, under CD. DF. Ax. 2.B.Z

But those lides are about the equal y EBA, FDC (Ax. 10. B. 1. &

Prep. 1). 2. Therefore, AB:CD=DF:BE. And fince

DF=CD&B E=M (Prep. 2). 3. AB:CDCD: M.

P.7. &11.B.S. Which was to be demonstrated.

[ocr errors]

B.2.

P.14. B.6.

« ForrigeFortsett »