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But the rectangle of DF, FA is equal to DG, because FA is equal to FG (II. Def. 2.); and the difference of the squares of EB, EA is equal to the square of AB (II. 14. Cor.); whence ng is equal to Ac.
And, if from each of these equals, the part DH, which is common to both, be taken away, the remainder AG will be equal to the remainder Hc.
But hc is the rectangle of AB, BH; for AB is equal to BC; and AG is the square of Ah; therefore the right line ab is divided in h so, that the rectangle of AB, BH is equal to the square of AH, whịch was to be done,
1. A radius of a circle, is a right line drawn from the centre to the circumference.
2. A diameter of a circle, is a right line drawn through the centre and terminated both ways by the circumference.
3. An arc of a circle, is any part of its periphery, or circumference.
4. The chord, or subtense, of an arc, is a right line joining the two extremities of that arc.
5. A semicircle, is a figure contained under any diame. ter and the part of the circumference cut off by that diameter,
6. A segment of a circle, is a figure contained under any arc and the chord of that arc.
7. A tangent to a circle, is a right line which passes through a point in the circumference without cutting it
8. Right lines, or chords, are said to be equally distant from the centre of a circle, when perpendiculars drawn to them from the centre are equal.
9. And the right line on which the greater perpendis cular falls, is said to be farther from the centre.
10. An angle in a segment, is that which is contained by two right lines, drawn from any point in the arc of the segment, to the two extremities of the chord of that arc,
11. One circle is said to touch another, when it passes through a point in its circumference without cutting it.
PRO P. I. PROBLEM.
To find the centre of a given circle,
Let ABC be the given circle; it is required to find its centre,
Draw any chord a B, and bisect it in D (1. 10.); and through the point o draw ce at right angles to AB (E., 11.), and biseet it in F: then will the point f be the centre of the circle.
For if it be not, fome other point must be the centre, either in the line Ec, or out of it,
But it cannot be any other point in the line ec, for if it were, two lines drawn from the centre of the circle to its circumference would be unequal, which is absurd.
Neither can it be any point out of that line; for if it can, let G be that point; and join GA, GD and GB.
Then, because GẠ is equal to GB (1. Def. 13.), AD to DB (by Const.), and GD common to each of the triangles AGD, BGD, the angle ADG will be equal to the angle BDG (I. 7.)
But when one line falls upon another, and makes the adjacent angles equal, those angles are, each of them, right angles (I. Def. 8 and 9.)
The angle AdG, therefore, is equal to the angle ADC (I. 8.), the whole to the part, which is absurd; consequently no point but y can be the centre of the circle.
Q. E. D. COROLL. If any chord of a circle be bisected, a right line drawn through that point, perpendicular to the chord, will pass through the centre of the circle.
PROP. II. THEOREM.
If any two points be taken in the circumference of a circle, the chord, or right line which joins them, will fall wholly within the circle.
Let ABE be a circle, and A, B any two points in the circumference; then will the right line AB, which joins those points, fall wholly within the circle.
For find' c, the centre of the circle ABE (III.1.), and join C, A, C, B; and through any point D, in AB, draw the right line ce, cutting the circumference in e.
Then, because ca is equal to CB (I. Def. 13.), the angle cab will be equal to the angle CBA (I. 5.)
And, since the outward angle cds of the triangle ACD, is greater than the inward opposite angle CAB (I. 16.), it will also be greater than the angle CBA.