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it is required to draw a straight line from A which shall touch Book III. the circle. \_^^J Find - the center E of the circle, and join AE; and from * 1: 3. the center E, at the distance EA, describe the circle AFG ; from the point D draw b DF at right angles to EA, and join " " " EBF, AB; AB touches the circle BCD. Because E is the center of the circles BCD, AFG, EA is equal to EF: And ED to EB; therefore the two fides AE, EB are equal to the two FE, ED, and they contain the angle at E common to the two triangles AFB, FED ; therefore the base DF is equal to the base AB, and the triangle FED to the triangle AEB, and the other angles to the other angles" : Therefore the c 4: i. angle EBA is equal to the angle EDF : But EDF is a right angle, wherefore EBA is a right angle: And EB is drawn from the center; but a straight line drawn from the extremity of a diameter, at right angles to it, touches the circle 4: Therefore d Cor. 16.3. AB touches the circle; and it is drawn from the given point A. Which was to be done. But, if the given point be in the circumference of the circle, as the point D, draw DE to the center E, and DF at right angles to DE; DF touches the circle".

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Fa straight line touches a circle, the straight line drawn from the center to the point of contact, shall be perpendicular to the line touching the circle.

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Bock III. Therefore FC is greater than FG ;
Ur-Q but FC is equal to FB; therefore

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F a straight line touches a circle, and from the point * of contact a straight line be drawn at right angles to the touching line, the center of the circle shall be in that line.

Let the straight line DE touch the circle ABC in C, and from C let CA be drawn at right angles to DE; the center of the circle is in CA.

For, if not, let F be the center, if possible, and join CF: Because DE touches the circle ABC, and FC is drawn from the center to A the point of contact, FC is perpendicular * to DE ; therefore FCE is a right angle : But ACE is also a right angle; therefore the angle FCE is equal to the angle ACE, the less to B the greater, which is impossible: Wherefore F is not the center of the circle ABC : In the same manner, it may be shewn, that no other point D C E which is not in CA, is the center; that is, the center is in CA. Therefore, if a straight line, &c.

Q. E. D.
P R O P. XX. T H E O R.
Th; angle at the center of a circle is double of the

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that is, upon the same part of the circumference.

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Let ABC be a circle, and BEC an angle at the center, and Book iii. BAC an angle at the circumference, which have the same cir- \,-Y“J cumference BC for their base ; the angle A. BEC is double of the angle BAC.

First, Let E the center of the circle be within the angle BAC, and join AE, and produce it to F: Because EA is equal

to EB, the angle EAB is equal * to a 5. r. the angle EBA ; therefore the angles EAB, EBA are double of the angle EAB; b 32. I. but the angle BEF is equal * to the angles B C

EAB, EBA ; therefore also the angle F
BEF is double of the angle EAB : For the
same reason, the angle FEC is double of
the angle EAC : Therefore the whole angle BEC is double of
the whole angle BAC.
Again, Let E the center of the -
circle be without the angle BDC, and A
join DE and produce it to G. It
may be demonstrated, as in the first
case, that the angle GEC is double
of the angle GDC, and that GEB a
part of the first is double of GDB a
part of the other; therefore the re-
maining angle BEC is double of the
remaining angle BDC. Therefore
the angle at the center, &c. Q. E. D.

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HE angles in the same segment of a circle are e-Sce N. qual to one another.

Let ABCD be a circle, and BAD, BED angles in the same segment BAED : The angles BAD, BED are equal to one another.

Take F the center of the circle ABCD: And, first, let the segment BAED be greater than a semicircle, and join BF, FI): And because the B angle BFD is at the center, and the angle BAD at the circumference, C and that they have the same part of

F 3 the

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a 32. I.

b 21.3,

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To opposite angles of any quadrilateral figure described in a circle, are together equal to two right

angles.

Let ABCD be a quadrilateral figure in the circle ABCD . any two of its opposite angles are together equal to two right angles.

Join AC, BD ; and because the three angles of every triangle are equal" to two right angles, the three angles of the triangle CAB, viz. the angles CAB, ABC, BCA are equal to two right angles: But the angle CAB D is equal" to the angle CDB, because they are in the same segment BADC; NS C and the angle ACB is equal to the angle ADB, because they are in the fame segment ADCB: Therefore the whole angle ADC is equal to the A& R angles CAB, ACB : To each of these equals add the angle ABC; therefore the angles ABC, CAB, BCA are equal to the angles ABC, ADC : But ABC, CAB, BCA are equal to two right angles; therefore also the angles ABC, ADC are equal to two right angles: In the same manner, the angles

BAD,

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IMIL AR segments of circles upon equal straight see N. lines, are equal to one another.

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