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232. The radius perpendicular to a chor the chord and the arc subtended by it.

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Let AB be the chord, and let the radius 0. pendicular to AB at M.

Το prove AM BM, and arc AS-arc BS. Proof. Draw OA and OB from O, the centre of t In the rt. A OAM and OBM

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(having the hypotenuse and a side of one equal to the hypo side of the other),

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(equal at the centre intercept equal arcs on the circum

233. COR. 1. The perpendicular erected at the m chord passes through the centre. For the centre is from the extremities of a chord, and is therefore pendicular erected at the middle of the chord.

234. COR. 2. The perpendicular erected at the r chord bisects the arc of the chord.

235. COR. 3. The locus of the middle points of a parallel chords is the diameter perpendicular to ther

236. In the same circle, or equal circles, equal chords are equally distant from the centre; AND

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Let AB and CF be equal chords of the circle ABFC.
To prove AB and CF equidistant from the centre O.
Proof. Draw OP1 to AB, OHL to CF, and join OA and OC.
OP and OH bisect AB and CF,

(a radius to a chord bisects it).

Hence, in the rt. A OPA and OHC

§ 232

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(having a side and hypotenuse of the one equal to a side and hypotenuse

of the other).

:. OP=OH.

.. AB and CF are equidistant from O.

CONVERSELY: Let OP

To prove

=

он.

ABCF

Proof. In the rt. A OPA and OHC

the radius OA = the radius OC, and OP=OH (by hyp.).

.. AOPA and OHC are equal.

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§ 161

Ax. 6.

237. In the same circle, or equal circle chords are unequal, they are unequally dist the centre, and the greater is at the less dis

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In the circle whose centre is 0, let the o and CD be unequal, and AB the greater; a and OF be perpendicular to AB and CD respe To prove OE < OF. Proof. Suppose AG drawn equal to CD, and OH

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(in the same © two equal chords are equidistant from the

Join EH.

OE and OH bisect AB and AG, respective (a radius to a chord bisects it).

Since, by hypothesis, AB is greater than CD or its AE, the half of AB, is greater than AH, the half of

.. the ZAHE is greater than the

AE

(the greater of two sides of a ▲ has the greater ▲ opposi Therefore, the ZOHE, the complement of the less than the OEH, the complement of the A ..OE < OH,

(the greater of two of a ▲ has the greater side opposi ..OE< OF, the equal of OH.

238. CONVERSELY: In the same circle, or equal circles, if two chords are unequally distant from the centre, they are unequal, and the chord at the less distance is the greater.

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In the circle whose centre is 0, let AB and CD be unequally distant from 0; and let OE perpendicular to AB be less than OF perpendicular to CD.

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Proof. Suppose AG drawn equal to CD, and OH 1 to AG.

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(in the same two equal chords are equidistant from the centre).

Hence, OE < OH.

Join EH.

In the ▲ OEH the ZOHE is less than the

OEH, § 158

(the greater of two sides of a ▲ has the greater ▲ opposite to it).

Therefore, the AHE, the complement of the ZOHE, is greater than the Z AEH, the complement of the

(the greater of two

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.. AE> AH,

OEH.

$159

of a ▲ has the greater side opposite to it).

But AE AB, and AH- AG.

=

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239. A straight line perpendicular to a its extremity is a tangent to the circle.

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Let MB be perpendicular to the radius OA MB tangent to the circle.

To prove

Proof. From O draw any other line to MB, as O

OH> OA,

(a is the shortest line from a point to a straight l

.. the point His without the circle. Hence, every point, except A, of the line MB is circle, and therefore MB is a tangent to the circle

240. COR. 1. A tangent to a circle is perpendic radius drawn to the point of contact. For, if MB to the circle at A, every point of MB, except A the circle. Hence, OA is the shortest line from Ot is therefore perpendicular to MB (§ 114); that is, pendicular to OA.

241. COR. 2. A perpendicular to a tangent at t contact passes through the centre of the circle. For perpendicular to a tangent at the point of contact, fore, by § 89, a perpendicular erected at the point coincides with this radius and passes through the c

242. COR. 3. A perpendicular let fall from the circle upon a tangent to the circle passes through

contact.

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