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.

Let the st. line AB be bisected in C and produced to D.

Then must

rect. AD, DB together with sq. on CB=sq. 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. BGCD and BH-BD;

.. HG=CB;

..rect. MG-rect. AL.

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.

Ax. 3.

II. A.

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 is equal to the rectangle contained by the sum and difference of those lines.

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.

On CB describe the sq. CEFB.

Then must difference between sqq. on AC, CD=rect. AD, DB.

I. 46.

Draw DG to CE, and from it cut off DH=DB.

I. 31.

I. 31.

Draw HLK to AD, and AK || to DH.

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.

PROPOSITION VII. THEOREM.

If a straight line be divided into any two parts, the squares on the whole line and on one of the parts are equal to twice the rectangle contained by the whole and that part together with the square on the other part.

Let AB be divided into any two parts in C.

Then must

sqq. on AB, BC=twice rect. AB, BC together with sq. on AC.

On AB describe the sq. ADEB.

From AD cut off AH=CB.

Draw CF || to AD and HGK || to AB.

Then HF=sq. on AC, and CK=sq. on CB.

Then sqq. on AB, BC=sum of AE and CK

I. 46.

I. 31.

.. sqq. on AB, BC=twice rect. AB, BC together with sq. on AC.

Q. E. D.

Ex. If straight lines be drawn from G to B and from G to D, shew that BGD is a straight line.

If a straight line be divided into any two parts, four times the rectangle contained by the whole line and one of the parts, together with the square on the other part, is equal to the square on the straight line which is made up of the whole and the first part.

On AD describe the sq. AEFD.

H

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

Produce AB to D, so that BD=BC.

Then must four times rect. AB, BC together with sq. on AC=sq. on AD.

From AE cut off AM and MX each=CB.

I. 46.

Through C, B draw CH, BL || to AE.

I. 31.

Through M, X draw MGKN, XPRO || to AD.

I. 31.

Now. XE=AC, and XP=AC, .. XH=sq. on AC.

Also AG=MP=PL=RF,

II. A.

and CK=GR=BN=K0;

II. A.

.. sum of these eight rectangles

=four times the sum of AG, CK

-four times AK

=four times rect. AB, BC.

Then four times rect. AB, BC and sq. on AC =sum of the eight rectangles and XH

=AEFD

=sq. on AD.

Q. E. D.

PROPOSITION IX. THEOREM.

If a straight line be divided into two equal, and also into two unequal parts, the squares on the two unequal parts are together double of the square on half the line and of the square on the line between the points of section.

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

sum of sqq. on AD, DB=twice sum of sqq. on AC, CD.

Draw CE-AC at rt. 4s to AB, and join EA, EB.
Draw DF at rt. 4s to AB, meeting EB in F.

4s to EC, and join AF.

Draw FG at rt.

Then ACE is a rt. 4,

.. sum of 8 AEC, EAC-a rt. ;

and AEC= ▲ EAC,

.. LAEC half a rt. 4.

So also 4 BEC and 4 EBC are each-half a rt. 4.

Hence AEF is a rt. 4.

GEF is half a rt. 4, and ▲ EGF is a rt. ≤ ;

EFG is half a rt. 4;

:. ▲ EFG= 4 GEF, and .. EG=GF.

So also

BFD is half a rt. 4, and BD=DF.

Now sum of sqq. on AD, DB

=sq. on AD together with sq. on DF

=

sq. on AF

=sq. on AE together with sq. on EF

I. 32.

I. A.

I. B. Cor.

I. 47.

I. 47.

=sqq. on AC, EC together with sqq. on EG, GF I. 47.

=

twice sq. on AC together with twice sq. on GF

=twice sq. on AC together with twice sq. on CD.

Q. E. D.