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the remaining fquare of AC will also be equal to the remaining 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 fquares of AC, co are equal to the fquare of OA (II. 14.), and the fquares of DF, FO to the fquare of OD.

But the fquare of OA is equal to the fquare of OD (II. 2.); therefore the squares of AC, co are equal to the fquares of DF, Fo.

And fince 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 remaining fquare of Fo; and confequently co is equal to FO (II. 3.), as was to be fhewn.

COROLL. If two right angled triangles, having equal hypotenuses, have two other fides also equal, the remaining fides will likewise be equal, and the triangles will be equal in all respects.


A diameter is the greateft right line that can be drawn in a circle, and, of the reft, that which is nearer the centre is greater than that which is more remote.

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Let ABCD be a circle, of which the diameter is Ad, and the centre o; 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 oв, oc, OG and OF.

Then, because OA is equal to OB (I. Def. 13.), and OD to OC, AD is equal to Oв and oc taken together.

But OB, OC, taken together, are greater than BC (I. 18.); therefore AD is also greater than BC.

Again, the fquares of OH, Hв are equal to the fquare of OB (II. 14.), and the fquares of OK, KF to the fquare of OF.

But the fquare of OB is equal to the fquare of oF (II. 2. )} whence the fquares of OH, HB are equal to the fquares of OK, KF.

And fince FG is farther from the centre than BC (by Hyp.), Ok will be greater than OH (III. Def. 9.), and the fquare of ok than the fquare of oH (II. 4.) G 3


The remaining fquare of HB, therefore, is greater than the remaining square of KF, and HB greater than KF (II. 4.)

But BC is the double of BH, and FG is the double of FK (III. 3.); confequently BC is also greater than FG. Q. E. D.


A right line drawn perpendicular to the diameter of a circle, at one of its extremities, is a tangent to the circle at that point.




Let ABC be a circle whofe 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 EF, cutting the circumference of the circle in c..

Then, fince the angle EBD is a right angle (by Hyp.), the angles BEF, EFB will be each of them less than a right angle (I. 28.)

And, because the greater fide of every triangle is oppofite to the greater angle (I. 17.), the fide EF is greater. than the fide EB, or its equal EC.

But fince EF is greater than EC, the point F will fall without the circle ABC; and the fame may be shewn of any other point in BD, except B.

The line BD, therefore, cannot cut the circle, but muft fall wholly without it, and be a tangent to it at the point B, as was to be fhewn.

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.)


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 diftance EA, defcribe 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, fince E is the centre of the circles FDC, GAB, EA is equal to EB, and ED to EC.

And, because the two fides EA, 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. I.

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 poffible, fome 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, fince the greater fide of every triangle is oppofite to the greater angle (I. 17.), the fide FB will be greater than the fide FG.

But FB is equal to FC; therefore FC will also be greater than FG, a part greater than the whole, which is impoffible.

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