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DEF.—If any point be taken in a straight line, the line is said to be divided internally or externally, according as the point is in the line or in the line produced. The segments of the line are, in either case, the distances from the extremities of the given line to the point of section. A line is equal to the sum of its segments when it is divided internally, to their difference when divided externally.

Euclid II., 5 and 6. ç D

с А

B A

and

Let AB be a straight line bisected at C and divided inter

nally or externally at D, then AD is the sum of AC and CD, and BD is their difference.

(Prop. A) AD. DB=the difference of the sqq. on AC and CD. .. Fig. 1, AD. DB + CD2 = AC?, and Fig. 2, AD. DB + AC2 = CD2.

:

These two propositions, which we class as corollaries to Prop. A, were enunciated by Euclid as follows :

COR. 1.If a straight line be bisected and cut unequally, the rectangle contained by the unequal segments, together with the square on the line between the points of section, equals the square on half the line.-(Eu. II. 5.)

COR. 2.-If a straight line be bisected and produced to any point, the rectangle contained by the whole line thus produced and the part of it produced, together with the square on half the line bisected, equals the square on the line made up of the half and the part produced.—(Eu. II. 6.)

9. Divide a given straight line internally into two parts so that the rectangle contained by the parts may be a maximum (i.e., the greatest possible).

10. How should a straight line be divided into two parts so that the sum of the squares on the parts may be a minimum (i.e., the least possible)

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PROP. 9 and 10. THEOR. If a straight line be bisected and cut unequally, internally or externally, the sum of the squares on the unequal segments will be double of the sum of the squares on half the line and on the line between the points of section. ç D

B
A

B
Let AB be bisected at C, and cut unequally at D.
To prove ADP + DB2 2 AC? + 2 CD.

AD? = AC + CD + 2 AC.CD (II. 4).
BD2 + 2 BC. CD BC+ CDo (II. 7).
Adding and observing that because BC = AC,
:: 2 BC. CD = 2 AC.CD, and BC2 AC?.

:. ADP + BD2 = 2 AC? + 2 CD. Q.E.D. Note.--The algebraic proposition is, (a + b)2 + (a - b)2 = 2 a2 + 2 12. It may be interesting to the student to demonstrate the above proposition, as Euclid does, in dependence only on Book I.; for this purpose an outline of the demonstration is here given :

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Given AB bisected at C, and cut unequally at D.
Draw CE | AB and · AC. Join AE and EB. Draw

DF || CE and FG || AB.
Since CA = CE and LACE is a L., .. _CEA and <CAE

each = } Lt. (I. 5 and 32). Also L CEB and 2 CBE each = } L:, ... AEB is a _ :: Again, .:: LEGF LGCD [, and _ GEF='} L'i ..<GFE = } [": = < GEF.

.. GF GE (1. 6), and in like manner DF = DB.

=

a

.

Again, ::: AE = AC? + CEP (1. 47) = 2 AC?, and EF2 EG2 + GF2 = 2 GF2 2 CD2. .:. 2 AC2 + 2 CD2 = AE? + EF2 = AF= ADP + DF2

AD2 + DB2. Q.E.D. DEF.—The Projection of a finite line upon another line is the part of this line intercepted between perpendiculars dropped upon it from the extremities of the finite tine.

Thus, from A and B let perpendiculars AE and BF be drawn to CD, produced if necessary; then EF is the projection of AB upon CD.

B
B
A

А

F
с
CE

CE FD

B

PROP. 12. THEOR. In an obtuse-angled triangle the square on the side opposite to the obtuse angle is greater than the squares on the sides which contain that angle by twice the rectangle contained by either of them and the projection of the other upon it. Let AB be opposite the obtuse angle

ACB. Draw AD 1 BC produced,
then CD is the projection of AC

A

E

D

on BC.

To prove AB > BC? + AC by

B 2 BC.CD. :: BDP = BC + CD + 2 BC.CD (II. 4), Add AD2 to each. :: BD+ AD= BC? + CD+ AD2 + 2 BC.CD.

:: (1. 47) BA?

BC2 + CA? + 2 BC. CD, :: BA? > BC? + CA' by 2 BC.CD.

Q.E.D.

PROP. 13. THEOR. In any triangle the square on a side opposite to an acute angle is less than the sum of the squares on the sides containing that angle by twice the rectangle contained by either of them and the projection of the other upon it.

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B Let AB be opposite the acute - ACB. Draw AD 1 BC. To prove ABP < BC? + AC2 by 2 BC. CD. :: BD2 + 2 BC. CD BC+ CDo (II. 7), Add AD2 to eacb. :: BD? + ADP + 2 BC.CD BC? + CD + ADP.

=

:: (1. 47) BA? + 2 BC.CD BC+ AC?,

:. BA’ is < BC? + AC? by 2 BC.CD.

Note.—The case when AB coincides with AD needs no special mention; it comes at once by I. 47.

Propositions 12 and 13 may be enunciated together thus :

The square on one side of a triangle differs from the squares on the other two sides by twice the rectangle contained by either of them and the projection of the other upon it.

11. If the square on one side of a triangle be less or greater than the sum of the squares on the other two sides, the angle opposite the first side shall be less or greater than a right angle.

PROP. B. THEOR.

The sum of the squares on two sides of a triangle is equal to twice the square on half the base and twice the square on the line joining the vertex with the middle point of the base. Let D be the middle pt. of BC.

A To prove AB? + AC2 2 BD2

+ 2 DA?
Draw AE I BC.
ABP BD? + DAP + 2 BD. DE B-

D E
(II. 12).
AC = CD? + DA’ – 2 CD.DE (II. 13).
Since BD

BD2 CD, and 2 BD . DE =
2 CD. DE,
:: AB + ACP 2 BD+ 2 DA?.

Q.E.D.

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CD,

12. The squares on the sides of a parallelogram are together equal to the squares on the two diagonals.

13. The sum of the squares on the sides of a quadrilateral is greater than the sum of the squares on the two diagonals by four times the square on the line joining the middle points of the diagonals.

14. Find the locus of the vertex of a triangle when the base is given and also the sum of the squares on the other two sides.

15. If a line be drawn through the centre of two concentric circles, meeting the circumferences in A, B, C, and D, and points P and Q be taken on the circumferences, then PB2 + PC2 shall be equal to QAP+QD2.

16. In a trapezium the squares on the two diagonals are together equal to the squares on the two non-parallel sides, together with twice the rectangle contained by the parallel sides.

17. The square on the base of an isosceles triangle is equal to twice the rectangle contained by either side and the projection of the base upon it.

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