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PROP. VII. THEOREM.
If two circles touch each other externally, the centres of the circles and the point of contact will be all in the saine right line.
Let the two circles beg, BDF touch each other externally at the point B; then will the centres of those circles and the point B, be in the fame right line.
For, let a be the centre of the circle beg, and draw the diameter GB, which produce till it cuts the circle BDF in F.
And, if the centre of the circle BDF be not in the line AF, let, if possible, some point c, out of that line, be the centre; and join C, A, C, B.
Then, fince A is the centre of the circle BEG, Ae is equal to AB (I. Def. 13.)
And because c is the centre of the circle BDF (by Hyp. ), cd is equal to CB (I. Def. 13.)
But AB, BC, together, are greater than ac (1. 18.); therefore AE, CB, together, are also greater than AC; which is absurd.
The point c, therefore, cannot be the centre of the circle BDF; and the same may be Thewn of any other point out of the line af.
Q. E. D. PRO P. VIII. THEOREM.
Any two chords in a circle, which are equally distant from the centre, are equal to each others and if they be equal to each other, they will be equally distant from the
Let Altd be a circle, whole centre is ó; then will any two chords AB, DE, which are equally distant from 0, be equal to each other.
For join the points A0, od, and let fall the perpendiculars oc, oF (I. 12.)
Then, since a right line, drawn from the centre of a circle, at right angles to any chord, bisects it (III. 3.); AC will be equal to CB, and DF to FE.
And, because the angles Aco, DFO are right angles, the squares of ac, co will be equal to the square of so (II. 14.), and the squares of DF, Fo to the square of Do.
But the square of Ao is equal to the square of OD (II. 2.); consequently the squares of ac, co will be equal to the squares of DF, FO.
And since oc is equal to oF (III. Def. 8.), the square of oc will be equal to the square of oF (II. 2.) ; whence
the remaining square of ac will also be equal to the re. maining square of DF; or ac equal to DF (II. 3.), and AB to De (I. Ax. 6.)
Again, let the chord AB' be equal to the chord DE; then will oc, of, or their distances from the centre, be equal to each other.
For the squares of ac, co are equal to the square of OA (II. 14.), and the squares of DF, fo to the square
But the square of oa is equal to the square of OD (II. 2.); therefore the squares of ac, co are equal to the squares of DF, Fo.
And since Ac is the half of AB (III. 3.), and df is the half of DE (III. 3.), the square of ac is equal to the square of DF (II. 2.)
The remaining square of co is, therefore, equal to the Temaining square of Fo; and consequently co is equal to FO (II. 3.), as was to be shewn.
COROLL. If two right angled triangles, having equal hypotenuses, have two other sides also equal, the remaining fides will likewise be equal, and the triangles will be equal in all respects.
PRO P. IX. THEOREM.
A diameter is the greatest right line that can be drawn in a circle, and, of the rest, that which is nearer the centre is greater than that which is more remote.
Let ABCD be a circle, of which the diameter is AD, and the centre 0; then if Bc be nearer the centre than FG, AD will be greater than bc, and BC than FG.
For draw oH, OK perpendicular to BC, FG (I. 12.), and join oß, oc, og and of.
Then, because of is equal to OB (I. Def. 13.), and OD to oc, ad is equal to OB and oc taken together,
But os, oc, taken together, are greater than BC (I.18.); therefore AD is also greater than Bc.
Again, the squares of oh, HB are equal to the square of OB (II. 14.), and the squares of ok, KF to the square of oF.
But the fquare of ob is equal to the square of oF (II. 2.)} whence the squares of oH, HB are equal to the squares of OK, KF.
And since FG is farther from the centre than BC .(by Hyp.), ok will be greater than oh (III. Def. 9.), and the square of ok than the square of oH (II. 4.)
The remaining square of hb, therefore, is greater than the remaining square of ķF, and HB greater than KF' (II. 4.)
But ßc is the double of bh, and FG is the double of FK (III. 3.); consequently bc is also greater than FG.
Q. E. D.
PRO P. X. THEOREM.
A right line drawn perpendicular to the diameter of a circle, at one of its extremi, ties, is a tangent to the circle at that point,
Let ABC be a circle whose centre is E, and diameter AB; then if DB be drawn perpendicular to AB, it will be a tangent to the circle at the point B.
For in BD take any point F, and draw £F, cutting the circumference of the circle in c.
Then, since the angle EBD is a right angle (by Hyp.), the angles bef, EFB will be each of them less tħan a right angle (1. 28.)
And, because the greater side of every triangle is opposite to the greater angle (I. 17.), the side Er is greater than the side EB, or its equal EC.
Bụt since er is greater ţhan Eç, the point F will fall without the circle ABC; and the same may be thewn of any other point in BD, except B.