and passing through O the centre of the given circle. Let P be the centre of this circle; it will also be the centre of the circle required.

Join PM, PN, PO. Since these lines are equal, and MQ, RN, OF are also equal by construction, .. PQ, PR, PF are also equal; and a circle described from the centre P at the distance of any one of them, will pass through the extremities of the other two, and touch the lines AB, CD, in Q and R; since the angles at those points are equal to the angles at M and N, and .. right angles; and it will also touch the circle EFG in F, since OP the line joining the centres passes through F

(47.) To describe a circle which shall touch a circle and straight line, both given in position, and have its centre also in a given straight line.

Let the circle whose centre is A, and the straight line BC be given in position; and let CD be the line, in which the centre of the required circle is to be. On BC let fall the perpendicular AB; and make BF

F

B

E

H

G

= AE; through F draw FG parallel to BC, meeting DC in G.

Join GA; and draw CH parallel to it, meeting the given circle in H, (if the problem be possible). Join AH, and let it meet DC in O. O is the centre of the circle required.

Let fall the perpendicular O1. Then (Eucl. vi. 2.)

HO: OC :: AH: GC :: FB: GC by construction, :: BD: DC, (Eucl. vi. 2.)

:: 10 OC, by sim. triangles;

.. HO= IO; and a circle described with the centre O, and radius OI or OH, will pass through the extremity of the other, and touch the line BC in I, and the circle in H; because the angles at I are right angles; and 40 the line joining the centres of the circles passes through H.

(48.) Through two given points within a given circle, to describe a circle, which shall bisect the circumference of the other.

Let A and B be the given points within the circle whose centre is 0. Join 40; and produce it indefinitely; and from O draw OC at right angles to it. Join AC; and draw CD at right

angles to it, meeting 40 produced in D; and through A, B, D describe a circle; it will bisect the other in the points E, and F.

..the rectangle AO, OD is equal to the square of OC, i. e. to the rectangle EO, OF; whence (Eucl. iii. 35.) EOF is a straight line; and since it passes through the centre of the circle ECF, it will be a diameter of that circle; .. the circumference ECF is equal to the circumference EGF, or the circumference of the given circle is bisected.

(49.) Through two given points without a given circle, to describe a circle, which shall cut off from the given one an arc equal to a given arc.

H

Let A and B be the given points, and CDE the given circle. Join AB; and bisect it in F; from F draw FG at right angles to AB; and from any point G in it, at the distance GA or GB, describe a circle ABD, cutting the given circle in C and D. DC; and produce it to meet BA in H. From H draw HIE (ii. 20.), so that IE may be equal to the chord of the given arc. Through A, B and E describe a circle; it will also pass through I, and cut off the arc required.

Join

For the rectangle HI, HE is equal to the rectangle HC, HD, and .:. also to the rectangle HA, HB, whence I is a point in the circle ABE.

(50.) To describe three circles of equal diameters, which shall touch each other.

Take any straight line AB, which bisect in D; and from the centres A and B, with the equal radii AD, BD describe two circles. Upon AB describe an equilateral triangle ABC, cutting the circles in E and F; and with the centre C,

B

G

H

and radius CE or CF, describe another circle; these circles touch each other as required.

Since AB AC, and AD is half of AB, .. AE, which

is equal to it, is half of AC, and .. AE= EC. In the

same manner CF=FB, .. the radii of the circles are all equal. And the circles touch each other in D, E, F, because the lines joining the centres pass through those points.

(51.) Every thing remaining as in the last proposition; to describe a circle which shall touch the three circles.

Bisect the angles CAB, CBA (see last Fig.) by the lines AO, BO meeting in 0. O is the centre of the circle required.

Join OC. Since the angle CAB=CBA, .. their halves are equal, or OAB=OBA, .:. OA=OB. Also since CA= AB, and AO is common, and the angle CAO

BAO, .. CO=OB; and the three lines OA, OB, OC are equal; parts of which GA, HB, C'I are equal; .. OG, OH, OI are equal; and a circle described from O as a centre, with a radius equal to any one of those lines, will pass through the extremities of the other two, and touch the circles in the points G, H, I; because the lines joining the centres pass through those points.

In nearly the same manner a circle may be described which shall touch the three circles on the opposite circumferences.

(52.) To determine how many equal circles may be placed round another circle of the same diameter, touching each other and the interior circle.

Let A be the centre of the interior circle, and AD its radius. Describe (vi. 50.) the circles DF, EF touching the circle DE, and each other. Then the angle at A being one third part of two, or one sixth part of four right

E

angles, subtends an arc ED equal to one sixth of the whole circumference. And the same being true of every other contiguous circle, the number of circles which can be described touching each other and the interior one will be six.

(53.) From a given rectangular parallelogram to cut off a gnomon, whose breadth shall be every where the same, and whose area shall be to that of the parallelogram in any given ratio.

A

K

E

G

F

B

L

Let AC be the given parallelogram. Produce AB to D making BD = BC. On AD describe a semicircle, and produce CB to E; and let the ratio of the part to be cut off, to the whole, be that of 1 n. Make BE : BF :: n : n−1; and take BG a mean proportional between BE and BF. Bisect AD in O; and with the centre 0, and radius OG, describe a semicircle HGI; AH= ID, will be the breadth of the gnomon.

Make BLBI, and draw HK, LK parallel to the sides of the parallelogram; then AC: HL in the ratio compounded of the ratios of AB BH and BD : BI,