(16.) From a given point without a circle, to draw a straight line cutting the circle, so that the rectangle contained by the part of it without and the part within the circle shall be equal to a given square.

B

Let A be the given point, and BCD the given circle. From A draw AB touching the circle; and on it as a diameter describe a semicircle, cutting the given circle in C. Join AC, and produce it to D. Now if the side of the given square be equal to BC, the problem is possible.

For the rectangle AC, AD is equal to the square of AB, i. e. to the squares of AC and BC; take away from each the square of AC,.. the rectangle AC, CD is equal to the square of BC.

(17.) From a given point in the circumference of a semicircle, to draw a straight line meeting the diameter, so that the difference between the squares of this line and a perpendicular to the diameter from the point of intersection may be equal to a given rectangle.

DBO

Let A be the given point in the circumference of the semicircle; from it draw AD perpendicular to the diameter. Take O the centre, and divide DO in B, so that the rectangle contained by 2OB, BD may be equal to the given rectangle. Join AB; and draw BC perpendicular to DB. AB, BC are the lines required.

For (Eucl. ii. 12.) the square of AB together with twice the rectangle OB, BD is equal to the difference of

the squares of OA and OB, i. e. to the square of BC; ..the difference between the squares of AB and BC is equal to twice the rectangle OB, BD, i. e. to the given rectangle.

(18.) From a given point to draw two lines to a third given in position, so that the rectangle contained by those lines may be equal to a given rectangle, and the difference of the angles which they make with that part of the third which is intercepted between them may be equal to a given angle.

H

B F

E

Let A be the given point, and BC the line given in position. From A draw AD perpendicular to BC; make the angle DAE equal to the given angle; and produce AE, till the rectangle DA, AE, is equal to the given rectangle. On AE as a diameter describe the circle AFG, cutting BC in F and G. Join AF, AG; they are the lines required.

Draw GH perpendicular to the diameter AE; then the arc HA is equal to the arc AG, and the angle AGH to AFG;. the angle HGF is equal to the difference of the angles AGF, AFG. Now the right-angled triangles AIK, KDG have the angles at K equal, .. the angle KAI KGD; but KAI was made equal to the given angle; the difference of the angles AFG, AGF is equal to the given angle. And (Eucl. vi. C.) the rectangle AF, AG is equal to the rectangle DA, AE, i. e. to the given rectangle.

(19.) Two points being given without a given circle; to determine a point in the circumference, from which lines drawn to the two given points shall contain· the greatest possible angle.

Let A and B be the given points, and EDF the given circle whose centre is 0. Describe a

circle through A, B, O. Join EF,

BA, and produce them to meet in G. From G draw GD touching the given circle in D. Through

E

B

D, A, B describe another circle; then since the square of GD is equal to the rectangle EG, GF, i. e. to the rectangle AG, GB, .. GD touches the circle ABD.

AD, DB. (ii. 2.)

Join

D is the point required, as is evident from

(20.) From the bisection of a given arc of a circle to draw a straight line such that the part of it intercepted between the chord of that and the opposite circumference shall be equal to a given straight line.

{

D

B

Ꮐ

E

It

Let DAE be the given arc of the circle ABC, bisected in A; AFC the diameter, and HI the given straight line. Produce HI to K, so that the rectangle HK, KI may be equal to the rectangle FA, AC. From A place in the circle AB=HK; AB is the line required.

Join BC; then the angle AFG being a right angle is equal to the angle ABC, and the angle at A is common,.. the triangles AGF, ABC are equiangular,

and AF AG :: AB : AC,

... the rectangle GA, AB is equal to the rectangle FA, AC, i. e. to the rectangle HK, KI. But AB = HK, ·. AG=KI, and consequently GB=HI.

(21.) To draw a straight line through a given point, so that the sum of the perpendiculars to it from two other given points may be equal to a given line.

E

B

F

HAC

Let A, B, C be the three given points, A being that through which the line is to be drawn. Join AC, and produce it, making AD=AC. Join BD, and on it describe a semicircle; in which place BE equal to the given line. Join DE; and through A draw FAG parallel to DE; it is the line required.

For let fall the perpendicular CG, and draw DH parallel to BE; then the triangles ACG, AHD being similar, and AC≈AD, .. CG=HD=FE, FD being a parallelogram; .. BF and CG together are equal to BE, i. e. to the given line; and FH being parallel to ED, BF is perpendicular to FG.

(22.) To draw a straight line through one of three points given in position; so that the rectangle contained by the perpendiculars let fall upon it from the other two may be equal to a given square.

Let A, B, C be the three given points, and A the point through which the line is to be drawn. Join AB,

E

B

AC; and draw CD parallel to BA, and take CE a third proportional to AB and a side of the given square. On AC describe a semicircle; and from E draw EF at right angles to CD, and meeting the semicircle in F. Join AF, and produce it; it is the line required.

D F

Join CF, which will be perpendicular to AD; and from B draw BG perpendicular to AG. Since CE is parallel to BA, and CF to BG, the triangles ABG, CEF will be similar,

.. AB BG :: CF: CE,

..the rectangle BG, CF, is equal to the rectangle AB, CE. But since the side of the given square is, by construction, a mean proportional between AB and CE, the rectangle AB, CE, is equal to the given square; .. the rectangle BG, CF is equal to the given square.

(23.) A given straight line being divided into two parts; to cut off a part which shall be a mean proportional between the two remaining segments.

Η

F

B

D

E

Let AB be divided into two parts in the point C; bisect CB in D, and draw DE perpendicular, and equal to AD; and through the points B, C, E describe a circle; produce ED to F. Join AE, and bisect EF in O; and from O draw OG parallel to AB, meeting AE in G; and since AD=DE, .. GO=OE, and G is a point in the circumference. From G draw GH perpendicular to AC; H is the point required.