II. I.

Let E be the Centre of the Circle, and join AE; then about the Centre E, with the Distance EA, defcribe the Circle A F G ; draw DF * at Right Angles to E A, and join E B F and A B. I fay, the Right Line A B is drawn from the Point A, touching the Circle BCD.

For, fince E is the Centre of the Circles B CD, AFG, the Line E A will be equal to E F, and ED to E B. Therefore the two Sides A E, E B, are equal. to the two Sides F E, ED, each to each; and they contain the common Angle E. Wherefore the Base DF is equal to the Bafe A B, and the Triangle DEF equal to the Triangle E B A, and the remaining Angles of the one equal to the remaining Angles of the other. And fo the Angle E B A is equal to the Angle E D F. But E D F is a Right Angle. Wherefore E B A is also a Right Angle, and E B is a Line drawn from the Centre; but a Right Line drawn. from the Extremity of the Diameter of a Circle at 1 Cor. 16. of Right Angles to it, touches the Circle. Wherefore, A B touches the Circle; which was to be done.

this.

12. I.

32. I.

PROPOSITION XVIII.

THEOREM.

If any Right Line touches a Circle, and from the
Centre to the Point of Contact a Right Line be
drawn; that Line will be perpendicular to the
Tangent.

LET any Right Line D E touch a Circle ABC
in the Point C, and let there be drawn the Right
Line F C from the Centre F. I fay, FC is perpen-
dicular to D E.

For, if it be not, let FG be drawn from the Centre F, perpendicular to D E.

Now, because the Angle F G C is a Right Angle, + Cor. 3. of the Angle GCF will be an acute Angle; and accordingly the Angle FGC is greater than the Angle FCG; but the greater Angle fubtends the greater Side. Therefore F C is greater than FG. But FC is equal to F B. Wherefore F B is greater than F G,

19. I.

a lefs

II. I.

Let E be the Centre of the Circle, and join AE; then about the Centre E, with the Distance E A, defcribe the Circle A F G; draw DF * at Right Angles to E A, and join E B F and A B. I fay, the Right Line A B is drawn from the Point A, touching the Circle B C D.

For, fince E is the Centre of the Circles B CD, AFG, the Line E A will be equal to E F, and ED to E B. Therefore the two Sides A E, E B, are equal. to the two Sides F E, ED, each to each; and they contain the common Angle E. Wherefore the Base + 4.1. DF is equal to the Bafe A B, and the Triangle DEF equal to the Triangle E B A, and the remaining Angles of the one equal to the remaining Angles of the other. And fo the Angle EBA is equal to the Angle ED F. But EDF is a Right Angle. Wherefore E B A is also a Right Angle, and E B is a Line drawn from the Centre; but a Right Line drawn. from the Extremity of the Diameter of a Circle at Cor. 16. of Right Angles to it, touches the Circle. Wherefore, A B touches the Circle; which was to be done.

tbis.

12. I.

32. I.

PROPOSITION XVIII.

THEOREM.

If any Right Line touches a Circle, and from the
Centre to the Point of Contact a Right Line be
drawn; that Line will be perpendicular to the
Tangent.

LET any Right Line D E touch a Circle ABC
in the Point C, and let there be drawn the Right
Line F C from the Centre F. I fay, FC is perpen-
dicular to D E.

For, if it be not, let FG be drawn from the Centre F, perpendicular to D E.

Now, because the Angle F G C is a Right Angle, +Cor. 3. of the Angle GCF will be † an acute Angle; and accordingly the Angle FGC is greater than the Angle FCG; but the greater Angle fubtends the greater Side. Therefore FC is greater than FG. But FC is equal to F B. Wherefore F B is greater than F G,

19. I.

a lefs