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.

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 CE to AB, and make CE=AC.

Join EA, EB and draw EF || to AD and DF || to CE. Then 48 FEB, EFD are together less than two rt. 4s, :. EB and FD will meet if produced towards B, D in some pt. G.

Then

Join AG.

ACE is a rt. 4,

=

.. 48 EAC, AEC together a rt. 4,

and EAC= L AEC,

.. LAEC half a rt. 4.

So also 4 BEC=half a rt. 4.

Hence DBG, which EBC, is half a rt. 4,

and

=

BGD is half a rt. 4;

:. BD=DG.

Again, ・・ ▲ FGE=half a rt. 4, and EFG is a rt. 4, .. L FEG=half a rt. 4, and EF=FG.

Then sum of sqq. on AD, DB

= sum of sqq. on AD, DG

=sq. on AG

=sq. on AE together with sq. on EG

I. A.

I. 47.

=sqq. on AC, EC together with sqq. on EF, FG 1. 47.
=twice sq. on AC together with twice sq. on EF
=twice sq. on AC together with twice sq. on CD. Q.E.D

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.

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

Q. E. F.

Thus AB is divided in H as was reqd.

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.

PROPOSITION XII. THEOREM.

In obtuse-angled triangles, if a perpendicular be arawn rom 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 ▲, having ACB obtuse. From A draw AD 1 to BC produced.

Then must sq. on AB be greater than sum of sqq. 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.

Add to each sq. on DA: then

II. 4.

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,

and sqq. on CD, DA=sq. on CA ;

I. 47.

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. 1. 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.

Ex. 2. If ABC be an equilateral triangle, and AD, BE be perpendiculars to the opposite sides intersecting in F; shew that the square on AB is equal to three times the square on AF.

PROPOSITION XIII. THEOREM.

In every triangle, the square on the side subtending any of the acute angles is less than the squares on the sides containing that angle, by twice the rectangle contained by either of these sides and the straight line intercepted between the perpendicular, let fall upon it from the opposite angle, and the acute angle.

Fig. 1.

Fig. 2.

B

B

C

ABC acute.

Let ABC be any ▲, having the

From A draw AD 1 to BC or BC produced.

Then must sq. on AC be less than the sum of sqq. on AB,

BC by twice rect. BC, BD.

For in fig. 1 BC is divided into two parts in D,

and in fig. 2 BD is divided into two parts in C';

.. in both cases

sum of sqq. on BC, BD=sum of twice rect. BC, BD and sq. on CD.

Add to each the sq. on DA, then

II. 7.

sum of sqq. on BC, BD, DA=sum of twice rect. BC, BD and sqq. on CD, DA;

.. sum of sqq. on BC, AB=sum of twice rect. BC, BD and sq. on AC;

I. 47. .. sq. on AC is less than sum of sqq. on AB, BC by twice rect. BC, BD.

The case, in which the AD coincides with AC, needs no proof.

Q.E.D.

Ex. 1. Prove that the sum of the squares on any two sides of a triangle is equal to twice the sum of the squares on half the base and on the line joining the vertical angle with the middle point of the base.

PROPOSITION XIV. PROBLEM.

To describe a square that shall be equal to a given rectilinear figure.

Let A be the given rectil. figure.
It is reqd. to describe a square that shall = A.
Describe the rectangular BCDE=A.
Then if BE=ED, the BCDE is a square,
and what was reqd. is done.

But if BE be not

I. 45.

ED, produce BE to F, so that EF=ED. Bisect BF in G; and with centre G and distance GB, describe the semicircle BHF.

Produce DE to H and join GH.

Then, BF is divided equally in G and unequally in E,

..rect. BE, EF together with sq. on GE

Ex. 1. Shew how to describe a rectangle equal to a given square, and having one of its sides equal to a given straight line.

Ex. 2. Divide a given straight line into two parts, so that the rectangle contained by them shall be equal to the square described upon a straight line, which is less than half the line divided.