and side EF, opposite equals in each, is common,
... AF FB.

NOTES.

1. Since the triangle EAB must be isosceles, we have only to demonstrate that

(i.) The straight line joining the vertex of an isosceles triangle to the mid-point of its base is perpendicular to the base.

(ii.) The perpendicular from the vertex of an isosceles triangle to the base bisects the base.

2. The words 'which does not pass through the centre' might obviously be omitted from the second part of the enunciation without affecting its truth. Hence

Every diameter of a circle is an axis of symmetry (see p. 23).

Ex. 270.-The mid-points of a set of parallel chords of a circle lie in a straight line.

Ex. 271.—If a trapezoid be inscribed in a circle it must be symmetrical about one of its medians.

In other words it must be an axe. (See pp. 103 and 106.)

Ex. 272. An axe can always have a circle described about it.

PROPOSITION 4.

If in a circle two straight lines cut each other which do not both pass through the centre they do not bisect each other.

Let ABCD be a O, and let AC, BD, two st. lines in it, cut each other in a pt. E which is not the centre; then AC, BD do not bisect each other.

A

B

For, if possible, let AE=EC and BE=ED.

Find the centre F of the O.

Neither AC nor BD can pass through F, or it would be

bisected at F instead of at E. Join FE.

Alternative Proof.-Let the st. lines AC, BD in the

ABCD bisect each other at E; then E shall be the

centre.

The st. lines through Er to AC, BD must each pass through

the centre;

.. the centre must be at E, the only pt. common to these

Lrs.

Ө

... if two st. lines in a bisect each other, they both pass through the centre.

.. if two st. lines in a O do not both pass through the centre, they do not bisect each other.

NOTE.

The student will perhaps have already noticed in Book I. pairs of Propositions connected with each other in the same way as the one enunciated as III. 4, and the one directly demonstrated in the Alternative Proof. The general relationship between such a pair of Propositions may be thus stated :

The two Propositions,

If A is B then C is D,

If C is not D then A is not B,

are each a necessary consequence of the other, i.e. if either is true the other must also be true.

Each of the above theorems is called the contrapositive of the other. (Syllabus.)

In many cases we can show the truth of a Proposition which Euclid proved indirectly by demonstrating the contrapositive theorem directly.

PROPOSITION 5.

If two circles cut one another they shall not have the same centre.

Let the two Os ABC, DCG cut each other at C, they shall not have the same centre.

A

D

B

G

Find the centre E of the ABC; join EC, and draw any st. line EFG, cutting the Os in F and G.

... Rad. EF=rad. EC,

.. EG is not equal to EC,

... E is not the centre of the DCG.

Ex. 273.-Write down the contrapositive of III. 5.

DEF.-Circles are said to touch one another which meet, but do not cut another.

If each of two circles which touch each other is outside the other, they are said to touch each other externally; if one is wholly within the other, they are said to touch each other internally.

PROPOSITION 6.

If two circles touch one another internally they shall not have the same centre.

Let the Os ABC, CDE touch one another internally at C, they shall not have the same centre.

Find F, the centre of the inner OCDE; join FC, draw FE any other radius of CDE, and produce it to cut ABC in B.

Rad. FE=rad. FC,

... FB is not equal to FC,

... F is not the centre of ABC.

Ex. 274.-Write down the contrapositive of III. 6.

DEF.—If two circles have the same centre, they are said to be concentric.

From III. 5 and III 6 we see that

If two circles have a common point, they cannot be concentric. Hence, contrapositively,

If two circles are concentric, they cannot have a common point.

Ex. 275.—A chord AD of the larger of two concentric circles cuts the smaller in B and C. Show that ABCD. Show also that AB and CD subtend equal angles at the centre.