PROPOSITION IV. THEOREM.

180. CONVERSELY: In the same circle, or equal circles, equal arcs subtend equal angles at the centre.

In the equal circles ABP and A'B' P' let arc RS =arc R' S'.

We are to prove ROS R' O'S'.

Apply ABP to OA'B' P',

so that the radius O R shall fall upon O' R'.

Then S, the extremity of arc RS,

will fall upon S', the extremity of arc R' S',
(for RSR S', by hyp.).

.. OS will coincide with O'S',
(their extremities being the same points).

§ 18

.. Z ROS will coincide with, and be equal to, R' O'S'.

Q. E. D.

PROPOSITION V. THEOREM.

181. In the same circle, or equal circles, equal arcs are

In the equal circles ABP and A'B' P' let arc RS arc R' S'.

..AROS =AR'O'S',

(two sides and the included

(equal arcs in equal subtend equal at the centre).

of the one being equal respectively to two sides

and the included of the other).

§ 106

.. chord RS

=

chord R' S',

(being homologous sides of equal ▲).

Q. E. D.

PROPOSITION VI. THEOREM.

182. CONVERSELY: In the same circle, or equal circles,

equal chords subtend equal arcs.

In the equal circles A B P and A'B' P', let chord RS

= chord R'S'.

We are to prove arc RS = arc R' S'.

Draw the radii O R, O S, O' R', and O'S'.

(three sides of the one being equal to three sides of the other).

..20=20',

(being homologous & of equal §).

... arc RS = arc R' S',

$ 179

(in the same O, or equal, equal at the centre intercept equal arcs on the

PROPOSITION VII. THEOREM.

183. The radius perpendicular to a chord bisects the chord and the arc subtended by it.

Let A B be the chord, and let the radius CS be perpendicular to AB at the point M.

(being radii of the same );

..A ACB is isosceles,

(the opposite sides being equal);

.. 1 CS bisects the base A B and the ≤ C,

$ 84

§ 113

(the drawn from the vertex to the base of an isosceles ▲ bisects the base and

(equal at the centre intercept equal arcs on the circumference).

Q. E. D.

184. COROLLARY. The perpendicular erected at the middle

of a chord passes through the centre of the circle, and bisects the arc of the chord.

185. In the same circle, or equal circles, equal chords are equally distant from the centre; and of two unequal chords the less is at the greater distance from the centre.

In the circle ABEC let the chord A B equal the chord
CF, and the chord CE be less than the chord C F.
Let OP, OH, and OK be is drawn to these chords
from the centre 0.

We are to prove OP=OH, and OH < OK.

(two rt. § are equal if they have a side and hypotenuse of the one equal to a side and hypotenuse of the other).

Again,

.. OP=OH,

(being homologous sides of equal ▲).

since CEC F,

the OK will intersect C F in some point, as m.

(a is the shortest distance from a point to a straight line).

... much more is OK > OH.

Ax. 8

§ 52

Q. E. D.