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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 ZROS= R' O'S'.

Apply O A B P to O A' B' P',

so that the radius 0 R shall fall upon OʻR'.

Then S, the extremity of arc RS,

will fall upon S', the extremity of arc R' S',

(for R S = R' S', by hyp.).

§ 18

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

.:.Z ROS will coincide with, and be equal to, < R' OS'.

Q. E. D.

PROPOSITION V. THEOREM. 181. In the same circle, or equal circles, equal arcs are subtended by equal chords.

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In the equal circles A BP and A' B' P' let arc RS

= arc R' S'.

We are to prove chord R S = chord R' S'.

Draw the radii O R, O S, O' R', and O'S'.
In the A ROS and R' O'S"
OR= ('R',

§ 176
(being radii of equal ©),
OS= 0'S',

$ 176 20=LO',

§ 180 (equal arcs in equal © subtend equal t at the centre). ..A ROS= A R' O'S',

§ 106 (two sides and the included L of the one being equal respectively to two sides

and the included L of the other).
.. chord R S = chord R' S',
(being homologous sides of equal A).

Q. E. D.

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

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

chord R' S.

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§ 108

AR' O'S',
(three sides of the one being equal to three sides of the other).

.:. Δ ROS

..20= 20,
(being homologous é of equal A).
.. arc RS
= arc R' S,

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

circumference).

Q. E. D.

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 C S be per

pendicular to A B at the point M.
We are to prove A M= BM, and arc A S = arc B S.

Draw C A and C B.

CA=CB,
(being radii of the same 0);
.:. A AC B is isosceles,

§ 84 (the opposite sides being equal); ..ICS bisects the base A B and the ZC, § 113 (the I drawn from the vertex to the base of an isosceles A bisects the base and

the Z at the vertex).

.. A M = BM.
Also,

since ZACS= Z BCS,
arc A S= arc S B,

$ 179 (equal ts 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.

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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.

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CH,

E
In the circle A B E C let the chord A B equal the chord

CF, and the chord C E 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= 0 H, and 0 H < O K.

Join 0 A and O C.
In the rt. A A O P and COH

ОА
(being radii of the same 0);
АР

§ 183
(being halves of equal chords) ;
.:. Δ Α OP
ДСОН,

$ 109 (two rt. S are equal if they have a side and hypotenuse of the one equal to

a side and hypotenuse of the other).

.:: OP

(being homologous sides of equal A).
Again,

since CE < CF,
the I OK will intersect C F in some point, as m.
Now
OK > Om.

Ax. 8
But
Om > OH,

$ 52 (a I is the shortest distance from a point to a straight line).

... much more is O K> 0 H.

OH,

Q. E. D.

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