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The line BD, therefore, cannot cut the circle, but must fall wholly without it, and be a tangent to it at the point B, as was to be shewn.
SCHOLIUM. A right line cannot touch a circle in more than one point, for if it met it in two points it would fall wholly within the circle (III. 2.)
PROP. XI. PROBLEM.
From a given point to draw a tangent to a given circle.
Let A be the given point, and FdC the given circle ; it is required from the point a to draw a tangent to the circle FDC.
Find E, the centre of the circle FDC (III. 1.), and join EA; and from the point E, at the distance EA, describe the circle GAB.
Through the point D, draw DB at right angles to EA (I. 11.), and join EB, AC; and AC will be the tangent required.
For, since E is the centre of the circles FDC, GAB, EA is equal to EB, and ED to EC.
And, because the two sides E A, EC, of the triangle EAC, are equal to the two fides EB, ED, of the triangle EBD, and the angle E common, the angle ECA will also be equal to the angle EDB (I. 4.)
But the angle EDB being a right angle, the angle ECA is also a right angle; therefore fince Ac is perpendicular to the diameter ec, it will touch the circle FDC, and be a tangent to it at the point c (III. 10.)
Q: E. 1.
PRO P. XII. THEOREM.
If a right line be a tangent to a circle, and another right line be drawn from the centre to the point of contact, it will be perpendi, cular to the tangent.
Let the right line De be a tangent to the circle ABC at the point B, and, from the centre F, draw the right line FB; then will FB be perpendicular to de.
For if it be not, let, if possible, some other right line FG be perpendicular to De.
Then, because the angle FGB is a right angle (by Hyp.) the angle FBG will be less than a right angle (I. 28.)
And, since the greater fide of every triangle is opposite to the greater angle (I. 17.), the side FB will be greater than the side FG.
But FB is equal to FC; therefore FC will also be greater than FG, a part greater than the whole, which is impoffible.
The line FG, therefore, cannot be perpendicular to DE ; and the fame may be demonstrated of any other line but FB; consequently FB is perpendicular to de.
Q. E, D.
PRO P. XIII. THEOREM.
If a right line be a tangent to a circle, and another right line be drawn at right angles to it, from the point of contact, it will pass through the centre of the circle.
Let the right line de be a tangent to the circle ACB at the point B; then if AB be drawn at right angles to De, from the point of contact B, it will pass through the centre of the circle.
For if it does not, let F, if poffible, be the centre of the circle ; and join FB.
Then, since de is a tangent to the circle, and FB is a right line drawn from the centre to the point of contact, the angle Fbe is a right angle (III. 12.)
But the angle ABE is also a right angle, by construction; whence the angle Fbe is equal to the angle ABE; the less to the greater, which is impossible.
The point F, therefore, is not the centre; and the fame may be shewn of any other point which is out of the line AB; consequently AB must pass through the centre of the circle, as was to be thewn.
PRO P. XIV. ' THEOREM.
An angle at the centre of a circle is double to that at the circumference, when both of them stand upon the same arc,
Let the angle bec be an angle at the centre of the circle ABC, and Bac an angle at the circumference, both standing upon the same arc BC; then will the angle bec be double the angle BAC.
First, let E, the centre of the circle, be within the angle Bac, and draw AE, which produce to F.
Then, because Ea is equal to EB, the angle EAB will be equal to the angle EBA (I. 5.)
And, because AEB is a triangle, the outward angle BEF will be equal to the two inward opposite angles EAB, EBA, taken together (I. 28.)
But since the angles EAB, EBA, are equal to each other, they are, together, double the angle EAB; whence the angle BEF is also double the angle EAB.
And, in the same manner it may be shewn, that the angle rec is double the angle EAC; confequently the whole angle bec will also be double the whole angle BAC.
Again, let E, the centre of the circle ABC, fall without the angle Bac, and join ae.
Then, since the angle bre,of the triangle EFB, is equal to the angle cca of the triangle CAF (I. 15,) the remaining angles BeF, FBE of the one, are, together, equal to the remaining angles FAÇ, FCA of the other (I. 28.).
But the angle Fbe is equal to the angle EAF (I. 5.), and the angle FCA, or ECA, to the angle EAC (I. 5.); therefore the angles BeF, EAF, are, together, equal to the angles FAC, EAC.
And, if the angle EAF, which is common, be taken away, the remaining angle ber or bec, will be equal to twice the angle fac, Or BAC, as was to be shewn.
PRO P. XV. THEOREM.
All angles in the same segment of a circle are equal to each other,
Let ABCD be a circle, and BAC, BDC any two angles in the same fegment BADC; then will the angles BAC, BDC be equal to each other.
For, first, let the segment BADC be greater than a semicircle, and having found the centre E, join BE and EC.