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Let AOB, ACB be angles at the centre and at the circumference which stand upon the same arc AB.

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First, let AC, one of the lines, containing ACB, pass

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Secondly, let neither AC nor BC pass through O. Join CO and produce it to D.

and

or

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Therefore whole (fig. 2) AOB = twice whole ACB.

rem. (fig. 3) AOB=twice rem. ACB.

When the arc ADB is greater than a semicircle, the angle AOB will be greater than two right angles.

We shall call it the convex angle AOB, and denote

it by AOB, to distinguish it from the angle AOB standing on the arc ACB

B

D

It is easily seen that every part of the proof given above applies directly to the case of a convex angle such as ДOB.

THEOREM XIV.

Angles in the same segment, or which stand upon the same arc, are equal.

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For each is half of the angle at the centre.

Thus ACB ADB = AEB.

=

THEOREM XV.

If the whole circumference be divided into two parts, ACB, ADB, the angles which stand upon these parts are supplementary; that is,

the angle ADB + the angle ACB

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Therefore ADB + ACB = † {AOB+AOB},

but AOB+A OB = 4R.

Therefore ADB +ACB = 2R.

COR. 1. The opposite angles of any quadrilateral figure inscribed in a circle are equal to 2R.

COR. 2. The angle on a semicircumference or in a semicircle is a right angle.

A

B A

B

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COR. 3. The angle on an arc less than a semicircumference is less than a right angle; and the angle on an arc greater than a semicircumference is greater than a right angle.

THEOREM XVI.

If a straight line EF touch a circle in the point B, and from B BA be drawn cutting the circle, the angle ABE shall be equal to the angle upon the arc ADB, and the angle ABF to that upon the arc ACB.

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Draw the normal BC, which will pass through the centre and divide the circumference into two equal parts. Then since the arc CA+ arc ADB = a semicircumference,

the angle on CA+ the angle on ÁDB=R; [II. 15. Cor. 2. CBA+ABE = R,

but

therefore CBA+ABE = the angle on CA + the angle on ADB,

and

Therefore

CBA is the angle on CA.

ABE = the angle on ADB.

Therefore also the supplements of these are equal, viz.

the angle ABF and the angle on ACB.

We give another proof.

Let ACB, AHB be angles on the arc ADB.

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Then

BHA+HAB = ABE+ HBF,

therefore ACB+HAB = ABE+ HBF.

[II. 14.

Now let H be moved along the circumference up to B,

AH, HB remaining always joined.

Then HAB is continually diminished,

also HBF is continually diminished,

and HK continually approaches the position of EF.

And when H coincides with B,

Therefore

HAB=0, and HBF=0.

ACB-ABE

1. Determine the locus of a point which is always equally distant from a given circumference. Discuss the several

cases.

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