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point B, and perpendicular to the same line KG; which is impossible hence, DE and FG meet in some point 0. Now, O O is on a perpendicu
lar to AB at its middle point, it is, therefore, equally distant from A and B (B. I., P. XVI.). For a like reason, O is equally distant from B and C. If, therefore, a circumference be de
scribed from O as a centre, with a radius equal to OA, it will pass through A, B, and C.
Again, O is the only point which is equally distant from A, B, and C for, DE contains all of the points which are equally distant from A and B; and FG all of the points which are equally distant from B and C; and consequently, their point of intersection O, is the only point. that is equally distant from A, B, and C hence, one circumference may be made to pass through these points, and but one; which was to be proved.
Cor. Two circumferences cannot intersect in more than two points; for, if they could intersect in three points, there would be two circumferences. passing through the same three points; which is impossible.
PROPOSITION VIII. THEOREM.`
In equal circles, equal chords are equally distant from the centres; and of two unequal chords, the less is at the greater distance from the centre.
1o. In the equal circles ACH and KLG, let the chords AC and KL be equal: then will they be equally distant from the centres.
For, let the circle KLG be placed upon ACH, so that the centre R shall fall upon the centre O, and the point
2o. Let AB be less than KL: then will it be at a
greater distance from the centre.
For, place the circle KLG upon ACH, so that R shall fall upon 0, and K upon A. Then, because the chord KL is greater than AB, the arc KSL is greater than AMB; and consequently, the point I will fall at a point C, beyond B, and the chord KL will take the direction AC.
Draw OD and OE, respectively perpendicular to AC and AB; then will OE be greater than OF (A. 8), and OF than OD (B. I., P. XV.): hence, OE is greater than OD. But, OE and OD are the distances of the two chords from the centre (B. I., P. XV., C. 1) : hence, the less chord is at the greater distance from the centre; which was to be proved.
Scholium. All the propositions relating to chords and arcs of equal circles, are also true for chords and arcs of one and the same circle. For, any circle may be regarded as made up of two equal circles, so placed, that they coincide in all their parts.
PROPOSITION IX. THEOREM.
If a straight line is perpendicular to a radius at its extremity, it will be tangent to the circle at that point; conversely, if a straight line is tangent to a circle at any point, it will be perpendicular to the radius drawn to that point.
1o. Let BD be perpendicular to the radius CA, at A then will it be tangent to the circle at A.
point A; it is, therefore, tangent to it at that point (D. 11); which was to be proved.
2o. Let BD be tangent to the circle at A: then will it be perpendicular to CA.
For, let E be any point of the tangent, except the point of contact, and draw CE Then, because BD is a tangent, E lies without the circle; and consequently, CE is greater than CA: hence, CA is shorter than any other line that can be drawn from C to BD; it is, therefore, perpendicular to BD (B. I., P. XV.,.C. 1) ; which was to be proved.
Cor. At a given point of a circumference, only one tan gent can be drawn. For, if two tangents could be drawn, they would both be perpendicular to the same radius at the same point; which is impossible (B. I., P. XIV.).
PROPOSITION X. THEOREM.
Two parallels intercept equal arcs of a circumference.
There may be three cases: both parallels may be secants; one may be a secant and the other a tangent; or, both may be tangents.
1o. Let the secants AB and DE be parallel: then will the intercepted arcs MN and PQ be equal.
For, draw the radius. CH perpendicular to the chord MP; it will also be perpendicular to NQ (B. I., P. XX., C. 1), and H will be the middle point of the arc MHP, and also of the arc NHQ hence, MN, which is the difference of HN and HM,
is equal to PQ, which is the difference of HQ and HP (A. 3); which was to be proved.
2o. Let the secant AB and tangent DE, be parallel: then will the intercepted arcs MH and PH be equal. For, draw the radius CH
to the point of contact H; it will be perpendicular to DE (P. IX.), and also to its parallel MP. But, because CH is perpendicular to MP, H is the middle point of the arc MHP (P. VI.): hence, MH and PH are equal; which was to be proved.
3o. Let the tangents DE and IL be parallel, and let H and K be their points of contact: then will the intercepted arcs HMK and HPK be equal.
For, draw the secant AB parallel to DE; then, from what has just been shown, we shall have HM equal to HP, and MK equal to PK: hence, HMK, which is the sum of HM and MK, is equal to HPK, which is the sum of HP and PK; which was to be proved.
PROPOSITION XI. THEOREM,
If two circumferences intersect each other, the points of intersection will be in a perpendicular to the line joining their centres, and at equal distances from it.
Let the circumferences, whose centres are
intersect at the points A and B: then will CD be perpendicular to AB, and AF will be equal to BF
For, the points A and B, being on the circumference whose centre is C, are equally distant from C; and being on
C and D,
the circumference whose centre is D, they are equally distant from D: hence, CD is perpendicular to AB at its middle point (B. I., P. XVI., C.); which was to be proved.