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.

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

Through H draw KLM || to AD.

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, CM, and LG

= sum of MG, 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 Prop. 5.

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.

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

Then rect. DF=rect. AL, ·: BF= AC, and BD=CL.
Also LGsq. 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

=sum of AK, HF, GE and CK

=sum of AK, HF and CE.

Now AK=rect. AB, BC,

CE=rect. AB, BC,

:: BE=AB;

HF=sq. on AC.

.. 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.

PROPOSITION VIII. THEOREM.

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.

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.

On AD describe the sq. AEFD.

From AE cut off AM and MX each=CB.
Through C, B draw CH, BL || to AE.

Through M, X draw MGKN, XPRO || to AD.
Now: XE=AC, and XP= AC, :. XH=sq. on AC.
Also AG=MP=PL=RF,

and CK=GR =BN=KO;

.. 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.

II. A.

II. A.

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 Cand 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.

Draw FG at rt. 4s to EC, and join AF.

Then

ACE is a rt. 4,

.. sum of 8 AEC, EAC=a rt. ;
and AEC= L EAC,

.. LAEC half a rt. 4.

So also BEC and 4 EBC are each

half a rt. 4.

Hence AEF is a rt.
GEF is half a rt. 4, and

=

4.

EGF is a rt. 4;

Also,
.. LEFG is half a rt. 4; .. 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.

=sqq. on AC, EC together with sqq. on EG, GF 1. 47.
=twice sq. on AC together with twice sq. on GF
=twice sq. on AC together with twice sq. on CD.

Q. E. D.

Ex. If in any triangle BAC a line AD be drawn bisecting BC in D, shew that the sum of the squares on AB, AC is equal to twice the sum of the squares on AD, BD.