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.

A

G

Join EC. Then the triangle ADC AFG :: AD1: 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

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.

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.

E

B

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.

(27.) To divide a given straight line into two such parts, that the square of one may be to the excess of a given rectangle above the square of the other, in a given

ratio.

Let AB be the given straight line. From B draw BC at right angles to AB, and make AB BC in the given ratio. Join AC. Find a mean proportional between the sides of the given rectangle ;

:

E

and with it as radius, and B as centre describe a circle cutting AC in D. Join BD, and draw DE parallel to BC; E is the point required.

For (Eucl. vi. 2.) AE: ED2 :: AB2: BC2. Now the square of ED is equal to the difference of the squares of BD and BE, i. e. to the difference of the given rectangle and the square of BE; .. the square of AE is to the difference between the given rectangle and the square of BE as AB BC, i. e. in the given ratio.

N.B. The given rectangle must not be less than the square of the perpendicular from B upon AC; and when BD is less than BC, there are two points E.

(28.) From any angle of a triangle, not isosceles about the angle, to draw a line without the triangle to the opposite side produced, which shall be a mean proportional between the segments of the side.

Let ABC be the triangle, and B the angle from which the mean proportional is to be drawn. About the triangle describe a circle, and to the point B

Y

D

B

draw a tangent BD meeting the side CA produced in D. BD is a mean proportional between AD and DC.

(Eucl. iii. 36.) the rectangle AD, DC is equal to the square of DB; and .. AD : DB :: DB : DC.

(29.) From the obtuse angle of any triangle, to draw a line within the triangle to the opposite side, which shall be a mean proportional between the segments of the side.

Let ABC be a triangle having the obtuse angle ABC. Describe a circle about it, and produce BA to D, making AD=AB. From D draw DE parallel

B

to AC, meeting the circle in E; join BE, cutting AC in F; BF will be a mean proportional between AF and FC.

For (Eucl. vi. 2.) BF: FE :: BA : AD,

and since BA=AD, :. BF=FE.

Now the rectangle AF, FC is equal to the rectangle BF, FE, i. e. to the square of BF;

.. AF: FB :: FB : FC.

(30.) From the common extremity of the diameters of two semicircles given in magnitude and position; to draw a line meeting the circumferences, so that the rectangle contained by the two chords may be equal to a given square.

Let AB, AC be the diameters drawn from A, and given in magnitude and position. With the centre A,

and radius equal to a side of the given square, describe a circle, cutting the lesser semicircle in D. Draw DE perpendicular to AC, and meeting the other semicircle in F. Join AF, and produce it to G; AG is the line required.

G

D

E

B

For joining GC, the triangles AGC, AFE are similar, .. AC AG :: AF: AE,

and.. the rectangle FA, AG is equal to the rectangle CA, AE, i. e. to the square of AD, which is equal to the given square.

(31.) To draw a line parallel to a given line, which shall be terminated by two others given in position, so as to form with them a triangle equal to a given rectilineal figure.

Let AB, AC be the lines given in position, AD the line to which it is required to draw a parallel. Describe a rectangular parallelogram AEFG equal to the given figure. Produce EF

E

H

B

to H; and take AK a mean proportional between DH and 2 EF; draw KC parallel to AD; KC is the line required.

For the angles DHA, CAK being equal, as also DAH, ACK, the triangles DAH, AKC are equiangular, and similar; whence

AKC

AHD :: AK: DH' :: 2 EF: DH:: 2 EF [× AE : DH × AE, double of the triangle rectangle EF, AE, i. e.

Now the rectangle DH, AE is AHD, .. AKC is equal to the to the given rectilineal figure.