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 urith 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=8q. on AD.
On AD describe the sq. 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. 3). Now ::: XE=AC, and XP=AC, .. XH=sq. on AC.

Also AG=MP=PL=RF,
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

Q. E. D.

II. A.

= sq. on AD.

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. 28 to AB, and join EA, EB.
Draw DF at rt. 28 to AB, meeting EB in F.
Draw FG at rt. Zs to EC, and join AF.

Then :: LACE is a rt. 1,
.. sum of 28 AEC, EAC=a rt. 2;

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

.. LAEC:=half a rt. L.
So also · BEC and · EBC are each=half a rt. 6.

Hence L AEF is a rt. L.
Also, :;: _GEF is half a rt. L, and . EGF is a rt. Z;

... EFG is half a rt. L;

... EFG= 2 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.

Q. E. D.

=sq. on AF

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.

I. 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 ::: 28 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,
... 28 EAC, AEC together=a rt. 2,

and :: LEAC= L AEC,

.:: LAEC=half a rt. L. So also zs BEC, EBC each=half a rt. L.

... AEB is a rt. L.
Also _ DBG, which = . EBC, is half a rt. 1,

and ... BGD is half a rt. 2;
.. BD=DG.

I. B. Cor. Again, : 2 FGE=half a rt. 7,

and – EFG is a rt. 1, I. 34. ... FEG=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

a

а

=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 other part.

F

=sq. on EF

Let AB be the given st. line.
On AB 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. Take from each the square on AE.

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,

I. 47.

PROPOSITION XII. TAEOREM.

In obtuse-angled triangles, if a perpendicular be drawn from either of the acute angles to the opposite side produced, the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle, by twice the rectangle contained by the side, upon which, when produced, the perpendicular falls, and the straight line intercepted without the triangle between the perpendicular and the obtuse angle.

1

Let ABC be an obtuse-angled 1, having LACB obtuse.

From A draw AD I to BC produced. Then must sq. on AB be greater than sum of $99. on BC, CA by twice rect. BC, CD.

For since BD is divided into two parts in C, sq. on BD=sum of sqq. on BC, CD, and twice rect. BC, CD.

II. 4. Add to each sq. on DA: then sum of sqq. on BD, DA=sum of sqq. on BC, CD, DA and twice rect. BC, CD.

Now sqq. on BD, DA=sq. on AB, I. 47.

and sqq. on CD, DA=sq. on CA ; I. 47. .. sq. on AB=sum of sqq. on BC, CA and twice rect. BC, CD.

.. sq. on AB is greater than sum of sqq. on BC, CA by twice rect. BC, CD.

Q. E. D. Ex. The squares on the diagonals of a trapezium are together equal to the squares on its two sides, which are not parallel, and twice the rectangle contained by the sides, which are parallel.