EF. The triangles ACE, DFC, having each a right angle, and the angles at C equal, are equiangular, whence CE: CF :: AC÷; CD, i. e. in the giving ratio.

(9.) From a given point between two indefinite right lines given in position, to draw a line which shall be terminated by the given lines, and bisected in the given point.

Let AB, AC be the given lines, meeting in A. From P the given point draw PD parallel to AC one of the lines, and make DE=DA. Join EP, and produce it to F; then will EF be bisected in P.

A

D

E

P

For since DP is parallel to AF, (Eucl. vi. 2.) EP: PF :: ED : DA, i. e. in a ratio of equality.

COR. If it be required to draw a line through P which shall be terminated by the given lines, and divided in any given ratio in P, draw PD parallel to AC, and take AD: DE in the given ratio, and draw EPF, it will be the line required.

(10.) From a given point without two indefinite right lines given in position; to draw a line such that the parts intercepted by the point and the lines may have a given ratio.

Let AB, AC be the given lines, and P the given point. Draw PD parallel to AC, and take AD: DE in the given ratio. Join PE, and produce it to F. Then PF: PE will be in the given ratio.

For the triangles PDE and AEF are similar, having the angles at E equal, as also the angles PDE, EAF, (Eucl. i. 39.)

E

F

.. FE EP :: AE: ED

and comp. PF: PE :: AD: DE, i.e. in the given ratio.

(11.) From a given point to draw a straight line, which shall cut off from lines containing a given angle, segments that shall have a given ratio.

Let ABC be the given angle, and P the given point, either without or within. In BA take any point A, and take AB: BC in the given ratio. Join AC, and from P draw PDE parallel to AC. PDE is the line required.

D

B

P

E

For since DE is parallel to AC, (Eucl. vi. 2.) DB : BE :: AB : BC, i. e. in the given ratio.

(12.) If from a given point any number of straight lines be drawn in a straight line given to position; to determine the locus of the points of section which divide them in a given ratio.

Let A be the given point, and BC the line given in position. From A draw any line AB, and divide it at E in the given ratio; through E draw EF parallel to BD; it is the locus required.

B D

FA

From A draw any other line AD meeting EF in F; then (Eucl. vi. 2.) AF : FD :: AE ; EB, i. e. in the given ratio. In the same manner any other line drawn from A to BD will be divided in the given ratio by EF, which therefore is the locus required.

(13.) A straight line being drawn parallel to one of the lines containing a given angle and produced to meet the other; through a given point within the angle, to draw a line cutting the other three, so that the part intercepted between the two parallel lines may have a given ratio to the part intercepted between the given point and the other line.

to DE or AB, and take BE: CF in the given ratio. Join FP and produce it to A; APF is the line required. For since DE and CP are parallel to AB,

AD: EB :: DF : EF :: PF : CF

...alt. AD: PF :: EB: CF i. e. in the given ratio.

(14.) Two parallel lines being given in position ;

to draw a third, such that, if from any point in it lines be drawn at given angles to the parallel lines, the intercepted parts may have a given ratio.

Let AB, CD be the given parallel lines; in AB take any point E; and draw EF, EG making angles equal to the given angles; produce EF, and take EH EG in the given ratio;

A

L

E

-B

K

F

D

G

H

produce FE to I so that FI: IE :: HE : EF; through I draw LI parallel to AB; it is the line required.

Draw IK parallel to EG; then the triangles IEK, EFG are equiangular,

.. IE : IK :: EF: EG

but FI IE :: HE: EF;

.. ex æqu. FI: IK :: HE : EG, i. e. in the given ratio; and IKE=EGF which is one of the given angles, and by construction IFG is equal to the other. Also lines drawn from any point in LI, making with AB and CD angles equal to the given angles will be parallel and equal to FI, IK, and .. in the given ratio.

(15.) If three straight lines drawn from the same point and in the same direction be in continued proportion, and from that point also a line equal to the mean proportional be inclined at any angle; the lines joining the extremity of this line and of the proportionals will contain equal angles.

:

Let AB AC :: AC : AD, and from A let AE be drawn equal to AC, inclined at any angle to AB; join EB, EC, ED; the angle BEC B is equal to the angle CED.

E

D

For since AB : AC :: AC: AD, and AE=AC;

:. AB : AE :: AE : AD, i.e. the sides about the angle A are proportional, and.. the triangles AEB, AED are similar, and the angle AED is equal to EBA. Also since CAAE, the angle AEC-ECA; but ECA is equal to the two BEC, EBC, (Eucl. i. 32.) .. also AEC is equal to the two BEC, EBC; of which DEA is equal to EBC; .. the remainder DEC is equal to the remainder BEC.

B

(13.) A straight line being drawn parallel to one of the lines containing a given angle and produced to meet the other; through a given point within the angle, to draw a line cutting the other three, so that the part intercepted between the two parallel lines may have a given ratio to the part intercepted between the given point and the other line.

Let ABC be the given angle, DE parallel to AB, and P the given point.

From P draw PC parallel

A

B

DA P

to DE or AB, and take BE: CF in the given ratio. Join FP and produce it to A; APF is the line required. For since DE and CP are parallel to AB,

AD: EB :: DF: EF:: PF: CF

..alt. AD PF :: EB: CF i. e. in the given ratio.

:

(14.) Two parallel lines being given in position; to draw a third, such that, if from any point in it lines be drawn at given angles to the parallel lines, the intercepted parts may have a given ratio.

Let AB, CD be the given parallel lines; in AB take any point E; and draw EF, EG making angles equal to the given angles; produce EF, an take EH EG in the given rat produce FE to I so