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PART. ENUN. and CONST.-Let ABCD be a trapezium, whose side AB is to CD. Produce CD to E, and make DE

= AB. Through D and B

draw DF and BG to A

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CE = CD + DE = CD + AB (Const.); and the trapezium

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BF DE- DG =

BGEH (Prop. XXXVI.); and

m

the ▲ FBD = ▲ BDG (Prop. XXXIV.); ., by addition, the trapezium ABCD = BDEH (Ax. 2) Wherefore, if two sides &c.

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=

-Q. E. D.

PROP. LL. THEOR.

ACEH.

GEN. ENUN.-The sum of the areas of any two parallelograms described on the two sides of a triangle, is equal to that of a parallelogram on the base, whose side is equal and parallel to the line drawn from the vertex of the triangle to the intersection of the sides of the former parallelograms produced to meet.

H

E

G

A

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PART. ENUN. and CONST.-Let ABC be a A, ABDE, ACFGms described upon its sides AB, AC. Produce DE, FG to meet in H. Join HA, and produce it to L, making KL = AH. Through B draw BM to KL (Prop. XXXI.), and complete the Om BMNC; then the BMNCms ABDE + ACFG. Produce MB, NC to o and P.

m

DEMONST.-1. Now, because оH is to BA, and Bo to AH, .. ABOH is a ".

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K

B

M

I

2. Also them ABDE =

ABOн, upon the same base

AB, and between the same | AB, DH, (Prop. XXXV.) base KL, and between the same || $

m

=

BMLK, upon =

HL, OB. (Prop. XXXVI.)

3. In like manner, □m ACFG = □" KLNC.

4. .' ms

= m BMNC.

ABDE + ACFG = □ms BMLK + KLNC (Ax.2)

Wherefore the sum of the areas &c.-Q. E. D.

PROP. XXXVII. THEOR.

GEN. ENUN.-Triangles upon the same base, and between the same parallels, are equal to one another.

E

A

PART. ENUN.-Let the ▲ ABC, DBC be the same base BC, and between the same || AD, BC; then ▲ ABC

=ADBC.

S

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upon

F

spectively to CA and BD, and meeting AD produced in E and F; .. the figures EBCA, DBCF are ms. (Prop. XXXIV. Def.)

DEMONST.-1. Now the □m EBCA =

DBCF upon the same base BC, same . (Prop. XXXV.)

m

and between the

2. But the ▲ ABC = m 10 EBCA, because diam. AB bisects it; and the DBC

=

□ DBCF, because diam. DC bisects it (Prop. XXXIV.); .'. ▲ ABC = ▲ DBC (AX. 7). Wherefore s &c.-Q. E. D.

Upon this Theorem depends the following:

PROP. MM. PROB.

To describe a triangle, which shall be equal to a given pentagon, and of the same altitude.

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PART. ENUN.-Let ABCDE be the gn. pentagon. It is required to describe a ▲ of area, and the same altitude. CONST.-Join AC, AD. Through B and E draw BF, EG respectively to AC and AD, and meeting CD produced in F and G. (Prop. XXXI.) Join af, ag. Then AFG is the ▲ required.

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3. To each side add the ▲ ACD; .. the pentagon ABCDE

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=

to the gn.

from

Wherefore a A AFG has been described pentagon ABCDE, and of the same altitude, viz. the

A upon CD.-Q. E. F.

PROP. XXXVIII. THEOR.

GEN. ENUN.-Triangles upon equal bases, and between the same parallels, are equal to

one another.

S

G

A

D

H

PART. ENUN.-Let the AS ABC, DEF be upon = bases BC, EF, and between the same || BF, AD; then ABC= A DEF. CONST.-Through B

and F draw BG, FH respectively to CA, ED,

B

CE

F

and meeting AD produced in G and H (Prop. XXXI.); then the figures GBCA, DEFH

ms. (Def.)

DEMONST.-1. Now the □m GBCA

DEFH, upon the

S

same || GH, BF.

2. But the

m

are

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base EF, and between the

(Prop. XXXVI.)

ABC = 1⁄2 □m GBCA, because

diam. AB bisects it; and the ▲ DEF = 1/ □ DEFH, because diam. DF bisects it (Prop. XXXIV.); .'. ▲^ abc = ^ def (Ax. 7).

Wherefore &c. &c.-Q. E. D.
The following Propositions may be added here :-

PROP. NN. PROB.

GEN. ENUN. To bisect a given triangle by a line drawn from one of its angles.

PART. ENUN.-Let ABC be the gn. A. It is required to bisect it by a line drawn from one of its 2s; for instance, the

BAC.

CONST.-Bisect the opposite

side BC in D (Prop. X.), and join AD. The line AD bisects the ▲ ABC.

DEMONST.-For the AADE ▲ ADC, because they are B upon bases BD, CD, and of the

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same altitude; i. e. between the same || s. (Prop. XXXVIII.) Wherefore the ▲ ABC has been bisected by the st. line AD drawn from the BAC.- -Q. E. F.

PROP. OO. PROB.

GEN. ENUN.-To bisect a given triangle by a line drawn from a given point in one of its sides.

PART. ENUN.-Let ABC be a ▲, D a gn. pt. in the side

AC; then it is required to bi

sect the ▲ ABC by a line drawn through D.

B

CONST.-Bisect BC in E (Prop. X.), and join AE, DE; through A draw AF to DE (Prop. XXXI.), and join DF; then DF bisects the ▲ ABC.

A

D

G

C

DEMONST.-1. Because AF is | to DE, the A ADE=Afde. (Prop. XXXVII.)

2. From each take the ▲ GED; .. ▲ AGD = AFGE (Ax. 3).

3. Also, because EC EB, the ▲ ACEA AEB (Prop. XXXVIII.); .'., by subtraction, the trapezium DGEC = trapezium ABFG (Ax. 3), and ▲ GFE = ▲ AGD.

4. .., by addition, the ▲ DFC = trapezium ABFD (Ax. 2). Wherefore the ▲ ABC has been bisected by the st. line FD drawn through the gn. pt. D.—Q. E. F.

PROP. PP. PROB.

GEN. ENUN. To find a point within a given triangle, from which lines drawn to the several angles will divide the triangle into three equal parts.

PART. ENUN.-Let ABC be the gn. A. It is required to find a pt. within it, from which lines drawn to the s A, B, C, shall divide it into three parts.

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CONST.-Bisect AB, BC, in D and E. B. (Prop. X.) Join CD, AE, intersecting in F. Then F is the pt. required. Join BF.

D

E

E

=

DEMONST.-1. Because AD DB, .. the ▲ ADC ▲ BDC, and the ▲ ADF = ▲ BDF (Prop. XXXVIII.); .'., by subtraction, ▲ AFC ▲ BFC (AX. 3).

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2. Again, because BE EC, .'. the AAEB = ▲ AEC, and the▲ BEF = ▲ CEF (Prop. XXXVIII.); .., by subtraction, the ▲ AFB = ▲ AFCA BFC (Ax. 1 and 3).

Wherefore a pt. F has been found within the ▲ ABC, from which the lines FA, FB, FC divide the ▲ into three pts.-Q. E. F.

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