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PROP. QQ. PROB.

GEN. ENUN.-To trisect a given triangle from a given point within it.

PART. ENUN.-Let ABC be the gn. A, D the gn. pt. within it. It is required to trisect ▲ ABC by lines drawn from the pt. D.

CONST.-Trisect the side BC in E and F (Prop. Z). Join DE, DF; and from a draw AG, AH respectively to them. (Prop. XXXI.) Join DG, DH, AD; then these three lines will trisect the ▲ ABC.

Join AE, AF.

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DEMONST.-1. Because AG is | to DE, the ▲ ADGA AEG, upon the same base AG, and between the same || $. (Prop. XXXVII.)

2. To each add the ▲ ABG; .. the trapezium ADGB = ▲ AEB (AX. 2).

3. In like manner, the trapezium ADHC = ▲ AFC; .. by addition, ADGB + ADHC = AEB + AFC (Ax. 2).

4. Subtract each from the ▲ ABC; .. the ▲ GDH = Δ AEF (AX. 3).

= A AEB =

5. Now the ▲ AEF A AFC (Prop. XXXVIII.) ; .. the ▲ GDH = trapezium ADGB = trapezium ADHC (Ax. 1).

Wherefore the A ABD has been trisected from a gn. pt. D within it.-Q. E. F.

PROP. RR. THEOR.

GEN. ENUN.-If from any point in the diameter of a parallelogram straight lines be drawn to the opposite angles, they will cut off equal triangles.

PART. ENUN. and CONST.-Let ABCD be a ", of which

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DFE (Prop. XXXVIII.) ; .., by addition, ▲ ACE =
DCE (AX. 2).

2. But the ▲ ABC = ▲ DBC (Prop. XXXIV.); . subtraction, ▲ ABE = ▲ DBE (Ax. 3).

Therefore, if from any pt. &c.-Q. E. D.

by

PROP. SS. PROB.

GEN. ENUN.-To bisect a trapezium by a line drawn from one of its angles.

PART. ENUN. and CONST.-Let ABCD be the gn. trape

zium, BAC the from

which it is required to
be bisected. Draw the
diagonals AD, BC. Bi-
sect BC, which is oppo-
site the BAC, in E.

(Prop. X.) Join AE,
ED; and through E draw
FEG to AD. (Prop. XXXI.)
bisects the trapezium.

H

B

D

Join AHG. Then AHG

=

DEMONST.-1. Because CE = EB,.. the ▲ AEC AEB, and the ▲ DEC = BED (Prop. XXXVIII.); .. the figure AEDC = figure AEDB (Ax. 2).

2. Also the ▲ AEG = ▲ DEG, upon the same base EG, and between same || $ EG, AD (Prop. XXXVII.). Take away the common part EHG; .. the ▲ AEHA DHG (AX. 3). 3. Hence the ▲ ACG = figure AEDC A DHG = figure figure AEDB Δ ΑΕΗ =

AEDC
AGDB.

Δ ΑΕΗ

=

trapezium

Wherefore the trapezium ABCD has been bisected by a st. line AG drawn through one of the 4 BAC.-Q. E. F.

PROP. XXXIX. THEOR.

PART. ENUN.-Equal triangles upon the same base, and upon the same side of it, are between the same parallels.

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CONST. and DEMONST. (Ad absurd.):

1. Join AD; and if AD be not || to BC, through a draw AE || to BC (Prop. XXXI.), and join Ec; then the ▲ ABC = ▲ EBC, upon the same base, and between the same || s. (Prop. XXXVII.)

2. But the ▲ ABC = ▲ DBC (Hyp.); the ▲ DBC = ▲ EBC, the > = the <, which is impossible.

3. ... AE is not || to BC; and, in like manner, no other line except AD is to it. Wherefore =AS upon &c.-Q. E. D.

Hence the following deduction :

PROP. TT. THEOR.

GEN. ENUN.-If two sides of a triangle be bisected, the straight line which joins the points of bisection is parallel to the third side.

PART. ENUN.-Let ABC be any A; bisect the sides AB,

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A

D

GEN. ENUN.-Equal triangles upon equal bases, in the same straight line, and towards the same parts, are between the same parallels. PART. ENUN.-Let the = ▲ ABC, DEF be upon = bases BC, EF, in the same st. line BF, and towards the same parts; then they are between the same || s.

B

E

F

CONST. and DEMONST. (Ad absurd.) :1. Join AD; and if it be not to BF, through a draw AG || to BF (Prop. XXXI.), and join GF; then the ABC =A GEF, upon the base EF, and between same || S. (Prop. XXXVIII.)

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2. But the ▲ ABC = ▲ DEF (Hyp.) the A DEF = Δ GEF (Ax. 1), the > = <, which is impossible.

2. Hence AG is not manner, no other line except AD is Wherefore equal $ &c.-Q. E. D. To this may be subjoined the following: :

to BF; and, in like

PROP. UU. THEOR.

to it.

GEN. ENUN.-Trapeziums upon the same base, having the sides opposite to the base equal, and lying between the same parallels, are equal to one another; and equal trapeziums upon the same base, and between the same parallels, have their sides opposite to the base equal.

8

PART. ENUN.-1. Let the trapeziums ABCD, EBCF be upon the same base BC, and between the same || AF, BC; and let the side AD side EF. Then the trapezium ABCD

=

=

=

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=

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▲ EBC, upon same base BC, and between the same (Prop. XXXVII.); and the ▲ ADB = A FEC, upon the bases AD, EF, and between same || 5. (Prop. XXXVIII.); .'., by addition, trapezium ABCD = trapezium EBCF. 2. Conversely, let the trapeziums be same; then the side AD = the side EF. DEMONST. (Ad abs.)

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and between the

1. For, since trapezium ABCD = trapezium EBCF, and, as before, the ▲ DBC = ▲ EBC, ..., by subtraction, ▲ ADB = A CEF.

2. Now, if AD be not = EF, let AD be the >; cut of GD = EF (Prop. III.); and join GB.

3. Then the ▲ BDG = A CEF, since they are upon = bases GD, EF, and between same s. (Prop. XXXVIII.)

4. But the ▲ ABD = ▲ CEF; .. ▲ BDG = ▲ ABD, the <= >, which is impossible.

5. .. AD is not unequal to EF, i. e. it is = to it. Wherefore trapeziums &c.-Q. E. D.

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