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. ADER.

I. 46.

From AD cut off AH=CB.

I. 31.

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, •: BK=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 są. AEFD.

I. 46.
From AE cut off AM and MX each=CB.
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,

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

II. A. II. A. 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.

1. A.

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. Zs to AB, and join EA, EB.
Draw DF at rt. Zs to AB, meeting EB in F.
Draw FG at rt. Ls to EC, and join AF.

Then :: LACE is a rt. 1,
.: sum of 28 AEC, EAC=a rt. 6;

I. 32. and ::: 1 AEC= _ EAC,

.: L AEC=half a rt. L.
So also BEC and 2 EBC are each=half a rt. L.

Hence AEF is a rt. L.
Also, ; 2GEF is half a rt. 1, and EGF is a rt. L;

... EFG is half a rt, L;

.: L EFG= GEF, and .. EG=GF. I. B. Cor. So also - BFD is half a rt. 1, and BD=DF. Now sum of sqq. on AD, DB =sq. on AD together with sq. on DF

I. 47. =sq. on AE together with sq. on EF

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.

a

a

=sq. on AF

Q. E. D.

PROPOSITION X. THEOREM. If a straight line be bisected and produced to any point, the square on the whole line thus produced and the square on the part of it produced are together double of the square on half the line bisected and of the square on the line made up of the half and the part produced.

a

...

Let the st. line AB be bisected in C and produced to D.
Then must
sum of sqq. on AD, BD=twice sum of sqq. on AC, CD.

Draw CEI to AB, and make CE=AC.
Join EA, EB and draw EF || to AD and DF || to CE.
Then :: %s FEB, EFD are together less than two rt. 28,

.. EB and FD will meet if produced towards B, D in some pt. G.

Join AG.
Then :: LACE is a rt. 1,
68 EAC, AEC together=a rt. 2,
and ::: LEAC= 1 AEC,

I. A. .:: L AEC=half a rt. L. So also zs BEC, EBC each=half a rt. 6.

.:: L AEB is a rt. b.
Also - DBG, which= . EBC, is half a rt. L,

and ... - BGD is half a rt. L;
.: BD=DG.

I. B. Cor. Again, :: 4 FGE=half a rt. 2, and EFG is a rt. 1, I. 34.

.. ZFEG=half a rt. 1, and EF=FG. I. B. Cor. Then sum of sqq. on AD, DB =sum of sqq. on AD, DG

I. 47. =sq. on AE together with sq. on EG

I. 47. =sqq. on AC, EC together with sqq. on EF, FG I. 47. =twice sq. on AC together with twice sq. on EF =twice sq. on AC together with twice sq. on CD. Q. E. D.

a

=sq. on AG

PROPOSITION XI. PROBLEM. To divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts shall be equal to the square on the olher part.

T

G

=sq. on EF

Let AB be the given st. line.
On A B descr. the sq. ADCB.

1. 46. Bisect AD in E and join EB.

I. 10.
Produce DA to F, making EF=EB.
On AF descr. the sq. AFGH.

I. 46. Then AB is divided in H so that rect. AB, BH=sq. on AH.

Produce GH to K. Then ::: DA is bisected in E and produced to F, .. rect. DF, FA together with sq. on AE

II. 6. =sq. on EB,

:: EB=EF, =sum of sqq. on AB, AE.

I. 47. Take from each the square on A E.

Then rect. DF, FA=sq. on AB. Ax. 3. Now FK=rect. DF, FA, :: FG=FA.

.. FK=AC. Take from each the common part AK.

Then FH=HC;
that is, sq. on AH=rect. AB, BH, .:: BC=AB.
Thus AB is divided in H as was reqd.

Q. E. F. Ex. Shew that the squares on the whole line and one of the parts are equal to three times the square on the other part.