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PROPOSITION III. THEOREM.

If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts is equal to the rectangle contained by the two parts together with the square on the aforesaid part.

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Let the st. line AB be divided into any two parts in C.
Then must
rect. AB, CB=sum of rect. AC, CB and sq. on CB.
On CB describe the sq. CDEB.

I. 46. From A draw AF || to CD, meeting ED produced in F.

Then AE=

=sum of AD and CE. Now AE=rect. AB, CB, :: BE=CB,

AD=rect. AC, CB, .: CD=CB,

CE=sq. on CB. :: rect. AB, CB=sum of rect. AC, CB and sq. cn CB.

Q. E. D.

NOTE. When a straight line is cut in a point, the distances of the point of section from the ends of the line are called the segments of the line. If a line AB be divided in C,

AC and CB are called the internal segments of AB. If a line AC be produced to B,

AB and CB are called the external segments of AC.

PROPOSITION IV. THEOREM.

If a straight line be divided into any two parts, the square on the whole line is equal to the squares on the two parts together with twice the rectangle contained by the parts.

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Let the st. line AB be divided into any two parts in C.

Then must sq. on AB=sum of sqq. on AC, CB and Iwice rect. AC, CB. On AB describe the sq. ADEB.

I. 46. From AD cut off AH=CB. Then HD=AC. Draw CG || to AD, and HK || to AB, meeting CG in F. Then ::: BK=AH, .: BK=CB,

Ax. Ι. .: BK, KF, FC, CB are all equal ; and KBC is a rt. Z; .: CK is the sq. on CB.

Def. xxx. Also HG=sq. on AC, ::HF and HD each=AC.

Now AE=sum of HG, CK, AF, FE, and AE=sq. on AB,

HG=sq. on AC,
CK=sq. on CB,
AF=rect. AC, CB, :: CF=CB,
FE=rect. AC, CB, :: FG=AC and FK=CB.
=sum of sqq. on AC, CB and twice rect. AC, CB.

Q. E. D.

.. sq. on AB=

Ex. In a triangle, whose vertical angle is a right angle, a straight line is drawn from the vertex perpendicular to the base. Shew that the rectangle, contained by the segments of the base, is equal to the square on the perpendicular.

PROPOSITION V. THEOREM.

If a straight line be divided into two equal parts and also into two unequal parts, the rectangle contained by the unequal parts, together with the square on the line between the points of section, is equal to the square on half the line.

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Let the st. line AB be divided equally in C and unequally in D. Then must

rect. AD, together with sq. on CD=sq. on CB.

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I. 46.

On CB describe the

sq.

CEFB. Draw DG || to CE, and from it cut off DH=DB. Draw HLK || to AD, and AK || to DH.

I. 31.

I. 31.

Then rect. DF=rect. AL,

Also LG=sq. on CD,

::BF=AC, and BD=CL.
::LH=CD, and HG=CD.

Then rect. AD, DB together with sq. on CD

=AH together with LG
=sum of AL and CH and LG
=sum of DF and CH and LG
=CF

=sq. on CB.

Q. L. D.

PROPOSITION VI. THEOREM.

If a straight line be bisected and produced to any point, the rectangle contained by the whole line thus produced and the part of it produced, together with the square on half the line bisected, is equal to the square on the straight line which is made up of the half and the part produced.

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Let the st. line AB be bisected in C and produced to D.
Then must
rect. AD, DB together with sq. on CB=8q. on CD.
On CD describe the

sq.
CEFD.

I. 46.
Draw BG || to CE, and cut off BH=BD.

I. 31. Through H draw KLM || to AD

I. 31. Through A draw AK || to CE.

Now ::: BG=CD and BH=BD;

.. HG=CB;
.. rect. MG=rect, AL.

Ax. 3. II. A.

Then rect. AD, DB together with sq. on CB

=sum of AM and LG
= sum of AL and CM and LG
=sum of MG and CM and LG
CF

sq. on CD.

Q. E. D.

NOTE. We here give the proof of an important theorem. which is usually placed as a corollary to Proposition V.

PROPOSITION B. THEOREM.

The difference between the squares on any two straight lines 28 equal to the rectangle contained by the sum and difference of those es.

D B

K

E

G
Let AC, CD be two st. lines, of which AC is the greater,
and let them be placed so as to form one st. line AD.

Produce AD to B, making CB=AC.
Then AD=the sum of the lines AC, CD,

and DB=the difference of the lines AC, CD. Then must difference between sqq. on AC, CD=rect. AD, DB. On CB describe the

sq.
CEFB.

I. 46. Draw DG 11 to CE, and from it cut off DH=DB. I. 31. Draw HLK || to AD, and AK || to DH.

I. 31. Then rect. DF=rect. AL, ::BF=AC, and BD=CL.

Also LG=sq. on CD, :::LH=CD, and HG=CD.
Then difference between sqq. on AC, CD

=difference between sqq. on CB, CD
=sum of CH and DF
=sum of CH and AL
=AH
=rect. AD, DH
=rect. AD, DB.

Q. E. D. Ex. Shew that Propositions V. and VI. might be deduced from this Proposition.

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