Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

PROPOSITION XLVIII. THEOREM.

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

B

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

Then must 2 BAC be a rt. angle. From pt. A draw ADI to AC.

I. 11. Make AD=AB, and join DC. Then

:: AD=AB, .. sq. on AD=sq. on AB; I. 46, Ex. 2.

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; I. 47. and, by hypothesis, sq. on BC=sum of sqq. on AB, AC; .. sq. on DC=sq. on BC; ..DCSBC.

I. 46, Ex. 3. Then in As ABC, ADC, :: AB=AD, and AC is common, and BC=DC, ...BAC= DAC;

I. c. and _DAC is a rt. angle, by construction ; ...BAC is a rt. angle.

Q. E. D.

a

BOOK II.

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.

D

B

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.

a

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

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

E

H

D

F

B

G 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, : . EFG- LABC,

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

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.

[merged small][ocr errors][ocr errors][ocr errors][merged small][merged small]
[ocr errors]
[ocr errors]

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.

I. 31.
Through E, F, and D draw EK, FL, DM || to CH.
Then EK and FL, being each=CH, are each=AB.

Now CM=sum of CK and EL and FM.
And CM=rect. AB, CD, :: CH=AB,

CK=rect. AB, CE, ::: CH=AB,
EL=rect. AB, EF, :: EK=AB,

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

Q. E, D.

[ocr errors]

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 the 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.

[merged small][merged small][ocr errors]

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.

I. 46.

On AB describe the sq. ADEB.

Through C draw CF li to AD.
Then AE=su

=sum of AF and CE.

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.

PROPOSITION III. THEOREM.

If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts is equal to the rectangle contained by the two parts together with the square on the aforesaid part.

[merged small][merged small][ocr errors]

Let the st. line AB be divided into any two parts in C.
Then must
rect. AB, CB=sum of rect. AC, CB and sq. on CB.
On CB describe the
sq. CDEB.

I. 46. From A draw AF || to CD, meeting ED produced in F.

Then AE=sum of AD and CE. Now AE=rect. AB, CB, :: BE=CB,

AD=rect. AC, CB, ::: CD=CB,

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

Q. E. D.

NOTE. When a straight line is cut in a point, the distances of the point of section from the ends of the line are called the segments of the line. If a line AB be divided in C,

AC and CB are called the internal segments of AB.
If a line AC be produced to B,

AB and CB are called the external segments of AC.

« ΠροηγούμενηΣυνέχεια »