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and, because EA is equal to EC,

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the angle EAC is equal to the angle ECA;

therefore the whole angle BAC is equal to the two angles, ABC, ACB.

[Axiom 2.

But FAC, the exterior angle of the triangle ABC, is equal to the two angles ABC, ACB; [I. 32. therefore the angle BAC is equal to the angle FAC, [Ax. 1. and therefore each of them is a right angle. [I. Def. 10. Therefore the angle in a semicircle BAC is a right angle.

And because the two angles ABC, BAC, of the triangle ABC, are together less than two right angles,

and that BAC has been shewn to be a right angle,

[I. 17.

therefore the angle ABC is less than a right angle. Therefore the angle in a segment ABC, greater than a semicircle, is less than a right angle.

And because ABCD is a quadrilateral figure in a circle, any two of its opposite angles are together equal to two right angles;

[III. 22. therefore the angles ABC, ADC are together equal to two right angles.

But the angle ABC has been shewn to be less than a right angle;

therefore the angle ADC is greater than a right angle. Therefore the angle in a segment ADC, less than a semicircle, is greater than a right angle.

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COROLLARY. From the demonstration it is manifest that if one angle of a triangle be equal to the other two, it is a right angle.

For the angle adjacent to it is equal to the same two angles;

[I. 32.

and when the adjacent angles are equal, they are right angles. [I. Definition 10.

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If a straight line touch a circle, and from the point of contact a straight line be drawn cutting the circle, the angles which this line makes with the line touching the circle shall be equal to the angles which are in the alternate segments of the circle.

Let the straight line EF touch the circle ABCD at the point B, and from the point B let the straight line BD be drawn, cutting the circle: the angles which BD makes with the touching line EF, shall be equal to the angles in the alternate segments of the circle; that is, the angle DBF shall be equal to the angle in the segment BAD, and the angle DBE shall be equal to the angle in the segment BCD.

From the point B draw BA at right angles to EF, [I. 11. and take any point C in the arc BD, and join AD, DC,

CB.

Then, because the straight line EF touches the circle ABCD at the point B, [Hyp. and BA is drawn at right angles to the touching line from the point of contact B,

E

therefore the centre of the circle is in BA.

B

[Construction. [III. 19.

Therefore the angle ADB, being in a semicircle, is a right angle.

[III. 31.

Therefore the other two angles BAD, ABD are equal to a right angle.

But ABF is also a right angle.

[I. 32

[Construction.

Therefore the angle ABF is equal to the angles BAD, ABD.

From each of these equals take away the common angle ABD;

therefore the remaining angle DBF is equal to the remaining angle BAD,

which is in the alternate segment of the circle.

[Axiom 3.

And because ABCD is a quadrilateral figure in a circle, the opposite angles BAD, BCD are together equal to two right angles.

[III. 22.

But the angles DBF, DBE are together equal to two right angles.

[I. 13. Therefore the angles DBF, DBE are together equal to the angles BAD, BČD.

And the angle DBF has been shewn equal to the angle BAD;

therefore the remaining angle DBE is equal to the remaining angle BCD,

which is in the alternate segment of the circle.

Wherefore, if a straight line &c. Q.E.D.

[Axiom 3.

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On a given straight line to describe a segment of a circle, containing an angle equal to a given rectilineal angle.

Let AB be the given straight line, and C the given rectilineal angle: it is required to describe, on the given straight line AB, a segment of a circle containing an angle equal to the angle C.

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Then the angle AHB

in a semicircle is equal to the right angle C.

[III. 31.

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therefore the base AG is equal to the base BG;

[I. 4.

and therefore the circle described from the centre G, at the distance GA, will pass through the point B.

Let this circle be described; and let it be AHB.

The segment AHB shall contain an angle equal to the given rectilineal angle C.

Because from the point A, the extremity of the diameter AE, AD is drawn at right angles to AE, [Construction. therefore AD touches the circle. [III. 16. Corollary.

And because AB is drawn from the point of contact A, the angle DAB is equal to the angle in the alternate segment AHB.

But the angle DAB is equal to the angle C.

Therefore the angle in the segment AHB is angle C.

[III. 32. [Constr.

equal to the [Axiom 1.

Wherefore, on the given straight line AB, the segment AHB of a circle has been described, containing an angle equal to the given angle C. Q.E.F.

PROPOSITION 34. PROBLEM.

From a given circle to cut off a segment containing an angle equal to a given rectilineal angle.

Let ABC be the given circle, and D the given rectilineal angle: it is required to cut off from the circle ABC a segment containing an angle equal to the angle D.

Draw the straight

line EF touching the circle ABC at the point B; [III. 17.

and at the point B,in the straight line BF, make the angle FBC equal to the angle D. [I. 23. The segment BAC shall contain an angle equal to the angle D.

Because the straight line EF touches the circle ABC, and BC is drawn from the point of contact B, [Constr. therefore the angle FBC is equal to the alternate segment BAC of the circle.

But the angle FBC is equal to the angle D.

angle in the

[III. 32. [Construction.

Therefore the angle in the segment BAC is equal to the angle D.

[Axiom 1. Wherefore, from the given circle ABC, the segment BAC has been cut off, containing an angle equal to the given angle D. Q.E.F.

PROPOSITION 35. THEOREM.

If two straight lines cut one another within a circle, the rectangle contained by the segments of one of them shall be equal to the rectangle contained by the segments of the other.

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