(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. Join AD, DB. D is the point required, as is evident from (ii. 62.)

(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.

JH

A

D

B

F

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 K HK, KI may be equal to the rectangle FA, AC. From A place in the circle AB=HK; AB is the line required.

E

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,

..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

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,

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 D
from E draw EF at right angles to

E

B

CD, and meeting the semicircle in F. Join AF, and produce it; it is the line required.

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.

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.

A

G

H

Ᏼ

C

D

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.

For HG, being perpendicular to AD, is perpendicular also to GO, and .. is a tangent at G; .. the square of HG is equal to the rectangle CH, HB. But since AD= DE, .. AH=HG, and consequently the square of AH is equal to the rectangle CH, HB; and AH is a mean proportional between the two remaining segments CH and HB.

(24.) To draw a straight line making a given angle with one of the sides of a given triangle, so that the triangle cut off may be to the whole in a given ratio. Let ABC be the given triangle;

make the angle ACD equal to the given angle which the cutting line is to make with AC. Produce AB to D; and make AE : AB in the ratio

of the part to be cut off to the whole. Take AF a mean proportional between AE and AD; draw FG parallel to CD; FG is the line required.

Join EC. Then the triangle ADC : AFG :: AD2 : AF2 :: AD : AE :: ACD : ACE, and .. AFG =ACE.

But ACE: ACB :: AE : AB,

:. AFG: ACB :: AE : AB, i. e. in the given ratio.

(25.) Between two given straight lines containing a given angle, to place a straight line of given length, and subtending that angle, so that the segment of the one of them adjacent to the angle may be to the segment of the other which is not adjacent, in the ratio of two given lines.

Let ED, EF be the lines given in length and position. Produce one of them FE, till EG: ED in the given ratio. Join DG; and with the centre E, and radius equal to the given line to be placed, describe a circle cutting DG

H

in H; join EH, and draw HI parallel to EF, and IC parallel to HE. IC is the line required.

For (Eucl. vi. 2.) HI : ID :: GE : ED,

and HC being a parallelogram, HI=EC,

.. EC ID :: GE ED, i. e. in the given ratio; and IC=EH=the given line.

(26.) From two given points to draw two lines to a point in a third, such that the difference of their squares may be equal to a given square.

A

Let A and B be the given points; join AB; and from A draw AE perpendicular to it, and equal to a side of the given square. Join BE, and bisect it in F; from F draw the perpendicular FG, meeting AB in G; and from G draw GD perpendicular to AB, meeting CD in D; join AD, DB; these are the lines required.

Join GE, it is equal to GB. And (iv. 30.) the difference between the squares of BD and AD is equal to the difference between the squares of BG and GA, i. e. between the squares of EG and GA, or it is equal to the square of AE, i. e. to the given square.