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Similarly, by joining AE, BK it may be shewn that
CL=sq. AK.

Now sq. on BC=sum of □ BL and □ CL,

sum of sq. BG and sq. AK, =sum of sqq. on BA and AC.

Q. E. D.

Ex. 1. Prove that the square, described upon the diagonal of any given square, is equal to twice the given square.

Ex. 2. Find a line, the square on which shall be equal to the sum of the squares on three given straight lines.

Ex. 3. If one angle of a triangle be equal to the sum of the other two, and one of the sides containing this angle being divided into four equal parts, the other contains three of those parts; remaining side of the triangle contains five such parts.

Ex. 4. The triangles ABC, DEF, having the angles ACB, DFE right angles, have also the sides AB, AC equal to DE, DF, each to each; shew that the triangles are equal in every respect.

NOTE. This Theorem has been already deduced as a Corollary from Prop. E, page 43.

Ex. 5. Divide a given straight line into two parts, so that the square on one part shall be double of the square on the other.

Ex. 6. If from one of the acute angles of a right-angled triangle a line be drawn to the opposite side, the squares on that side and on the line so drawn are together equal to the sum of the squares on the segment adjacent to the right angle and on the hypotenuse.

Ex. 7. In any triangle, if a line be drawn from the vertex at right angles to the base, the difference between the squares on the sides is equal to the difference between the squares on the segments of the base.

PROPOSITION XLVIII.

THEOREM.

If the square described upon one of the sides of a triangle be equal to the squares described upon the other two sides of it, the angle contained by those sides is a right angle.

B

Let the sq. on BC, a side of ▲ ABC, be equal to the sum of the sqq. on AB, AC.

Then must BAC be a rt. angle.

From pt. A draw AD1 to AC.

Then

Make AD=AB, and join DC.
:: AD=AB,

I. 46, Ex. 2.

.. sq. on AD=sq. on AB;
add to each sq. on AC.

then sum of sqq. on AD, AC=sum of sqq. on AB, AC.

But DAC is a rt. angle,

.. sq. on DC=sum of sqq. on AD, AC;
and, by hypothesis,

sq. on BC=sum of sqq. on AB, AC;
.'. sq. on DC=sq. on BC;
... DC BC.

Then in As ABC, ADC,

I. 11.

and DAC is a rt. angle, by construction;
BAC is a rt. angle.

..

I. 47.

I. 46, Ex. 3.

:: AB=AD, and AC is common, and BC=DC,
.. L BAC=1 DAC;

I. c.

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INTRODUCTORY REMARKS.

THE geometrical figure with which we are chiefly concerned in this book is the RECTANGLE. A rectangle is said to be contained by any two of its adjacent sides.

Thus if ABCD be a rectangle, it is said to be contained by AB, AD, or by any other pair of adjacent sides.

BOOK II.

We shall use the abbreviation rect. AB, AD to express the words "the rectangle contained by AB, AD."

We shall make frequent use of a Theorem (employed, but not demonstrated, by Euclid) which may be thus stated and proved :

PROPOSITION A. THEOREM.

If the adjacent sides of one rectangle be equal to the adjacent sides of another rectangle, each to each, the rectangles are equal

in area.

Let

A

B

ABCD, EFGH be two rectangles:
and let AB EF and BC=FG.

D

E

H

с

F

Then must rect. ABCD=rect. EFGH.

For if the rect. EFGH be applied to the rect. ABCD, SO that EF coincides with AB,

then FG will fall on BC,

G

EFG= 2 ABC,

and G will coincide with C, BC=FG.

Similarly it may be shewn that H will coincide with D, .. rect. EFGH coincides with and is therefore equal to rect.

ABCD.

Q. E. D.

77

B

PROPOSITION I. THEOREM.

If there be two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the rectangles contained by the undivided line and the several parts of the divided line.

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And CM=rect. AB, CD,

CK=rect. AB, CE,
EL=rect. AB, EF,
FM-rect. AB, FD,

L

M

G

Let AB and CD be two given st. lines,

and let CD be divided into any parts in E, F.

Then must rect. AB, CD=sum of rect. AB, CE and rect. AB, EF and rect. AB, FD.

From C draw CG 1 to CD, and in CG make CH=AB.

Through H draw HM || to CD.

Through E, F, and D draw EK, FL, DM || to CH.
Then EK and FL, being each=CII, are each=AB.
Now CM sum of CK and EL and FM.
·.· CH=AB,

·· CH=AB,
· EK=AB,

· FL=AB;

.. rect. AB, CD= sum of rect. AB, CE and rect. AB, EF and rect. AB, FD:

I. 31.

Q. E. D.

Ex. If two straight lines be each divided into any number of parts, the rectangle contained by the two lines is equal to the rectangles contained by all the parts of one taken separately with all the parts of the other.

PROPOSITION II. THEOREM.

If a straight line be divided into any two parts, the rectangles contained by the whole and each of the parts are together equal to the square on the whole line.

D

a B

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

Then must

sq. on AB=sum of rect. AB, AC and rect. AB, CB.

On AB describe the sq. ADEB

Through C draw CF || to AD.

Then AE sum of AF and CE.

=

I. 46.

I. 31.

Now AE is the sq. on AB,

AF=rect. AB, AC,

· AD=AB,

CE=rect. AB, CB,

·

BE=AB,

.. sq. on AB=sum cf rect. AB, AC and rect. AB, CB.

Q. E. D.

Ex. The square on a straight line is equal to four times the square on half the line.

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