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A

B

D

C

E

To draw a right Line AC

from a given Point A, touching
a given Circle DBC.

Draw a right Line DA from D the Centre of the given Circle to the given Point A, cutting the Periphery in B, and from the Center D defcribe another Circle AE paffing thro' A; and from B. draw a PerpendicuHar to AD, meeting the Circle AE in E: draw ED meeting the Circle BC in C, then a right Line drawn from A to C shall touch the Circle DBC.

3

For DB = DC, and DE a = DA; and the Ang. D is common: therefore the Ang. ACD = EBD; whence the right Line AC touches the Circle in C. Q. E. F. :

4

C

F

PROP. XVIII.

D

EG

If any right Line AB
touches a Circle FEDC,
and a right Line FE joins
the Centre F and Point of
Contact E; this Line
Shall be perpendicular to
the Tangent AB.

If you deny it, let
B some other right Line
FG, drawn from the

Center F be perpendicular to the Tangent, and

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C

F

B

If any right Line AB touches a Circle, and from C the Point of Contact a right Line CE be drawn at right Angles to that Tangent; the Center of the Circle shall be in the faid Line CE.

If you deny it, let the without CE in F, and draw FC from F to the Point of Contact. Now the B is da right Angle: and consequent-d 18. 3. to the right Angle ECB, by the Hypo

nich is absurd.

D

PROP. XX.

In a Circle DABC, the Ang. BDC at the Centre is the double of the Angle BAC at the Peri

phery, the fame Arch BC being

C the Base of the Angles.

11 ax.

9. ax.

4

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Draw the Diameter ADE.

The external Angle BDE=DAB+DBA = 2DAB. In like manner the Ang. EDC= 2DAC. Therefore, in the first Cafe, the whole Ang. BDC2BAC; but in the third Cafe, the remaining d Ang. BDC = 2BAC. Q. E. D.

PROP. ΧΧΙ.

In a Circle EDAC, the Angles DAC and DBC that are in the Same Segment, are equal to each other.

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* 20. 3.

Cafe 1. If the Segment DABC be greater than a Semicircle, draw ED, EC from the Center E. Then shall 2 Ang. A2 = E2 = 2B.

a

Q. E. D.

15. 1.

Cafe 2. If the Segment be not greater than a Semicircle, the Sum of the Angles of the Triangle ADF, are equal to the Sum of the Angles of the Triangle BCF. Take from both AFD By Cafe BFC, and ADB = ACB, and there remains

I.

PROP.

DAC=DBC. Q.E.D.

B

XXII.

The oppofite Angles ADC, ABC of quadrilateral FiCgures ABCD described in a Circle, are equal to two right Angles.

Draw AC, BD. The D Ang. ABC + BCA + BAC= 2 right Angles. But

A

* 32. Ι.

if* AB, one Side of a quadrila- * See the described in a Circle, be continued foll. Figure.

ernal Angle EBC shall be equal to
Angle ADC, which is opposite to
ngle adjacent to EBC, as is mani-
p. 13. 1. and ax. 3.
Circle cannot be described about a
because the opposite Angles of it
d, or are less than two Right An-

D

SCHOLIU M.

BE

In a quadrilateral Figure (ABCD) if the oppofite Angles A and C be equal; then a Circle may be described aCbout that quadrilateral Figure.

For a Circle may pass through any three angular Points B, C, D (as shall be by the 5 Prop. Lib. 4.) and I fay the pass thro' A. For if not, let it pass Then if you draw the right Lines BF, the Ang. C+ F = 2 right Ang. Therefore AF. which is abfurd.

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a

PROP. XXIII.

b

Two fimilar and unequal Segments ABC, ADC, of two Circles, cannot be fet upon the same right Line

AC on the Came Side there

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* Def. 10.

316. 1.

For if they are said to be similar, draw CB cutting the Circumference in B and D; and join AD, and AB. Now because the Segments are suppos'd to be similar, the Ang. ADC*= ABC. Which is absurd.

PROP. XXIV.

Similar Segments ABC, DEF of Circles, being upon equal right Lines AC, DF, are equal to each other.

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* 23. 3.

b

For if the Base AC be laid upon the Base DF, they will coincide, because AC = DF. Whence the Segment ABC will coincide with the Segment DEF, (for if not, it will fall within or without the fame,) and so the Segments will not be similar, (which is contrary to the Hyp.) or else partly within or partly without; and so it will cut it in three Points. Which is babfurd. Therefore the Segment ABC = DEF. Q. E. D.

10. 3.

€ 8. ax.

PROP.

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