Draw any other line AB, and make the angle BAF =DAE. Then the angle BAD-FAE and ADB= AEF, the triangles ABD, AEF are equiangular, whence AB: AF :: AD : AE, in the given ratio. The same may be proved of any other lines drawn from A and containing an angle equal to the given angle, and one of them terminated in BC.

(34.) If from two given points, straight lines be drawn, containing a given angle and from each of them segments be cut off, having a given ratio; and the extremities of the segments of the lines drawn from one of the points be in a straight line given in position; to determine the locus of the extremities of the segments of lines drawn from the other.

Let A and B be the given points, and CD the line given in position. From A to CD draw any line AE. Make the angle EAF the given angle, and AE:

=

C

B

M

F

L

H

/G

AF in the given ratio, and let FG be the locus of the points F (i. 33.) Draw BH equal and parallel to AF, and through H draw HI parallel to GF. It is the locus required.

Draw any lines AK, BK containing the angle at K =the given angle. Make the angle LAM=the given angle; AL: AM in the given ratio, and M is in the line GF. And since AF is parallel to BH, and FM to HN, and BK to AM (since the angles BKA, LAM are equal) and AFBH, .. the triangles BHN, AFM are similar and equal, .. AM=BN;

but AL

AM is equal to the given ratio,

.. also AL: BN is equal to the given ratio.

And the same may be proved of any other lines drawn in

the same manner.

SECT. II.

́ ́(1.) Ir a straight line be drawn to touch a circle, and be parallel to a chord; the point of contact will be the middle point of the arc cut off by that chord.

Let CD be drawn touching the cir- C cle ABE in the point E, and parallel· to the chord AB; E is the middle point of the arc AEB.

A

G

E

B

F

Join AE, EB. The angle BAE is equal to the alternate angle CEA, and therefore to the angle EBA in the alternate segment, whence AE = EB, and (Eucl. iii. 28.) the arc AE is equal to the arc EB. COR. 1. Parallel lines placed in a circle cut off equal parts of the circumference.

If FG be parallel to AB; the arc EF = EG, whence AG= BF.

COR. 2. The two straight lines in a circle, which join the extremities of two parallel chords are equal to each other. For if AB, FG be parallel, the arcs AG, BF are equal, therefore (Eucl. iii. 29.) the straight lines AG, BF are also equal.

(2.) If from a point without a circle, two straight lines be drawn to the concave part of the circumference, making equal angles with the line joining the same point and the centre, the parts of the lines which are intercepted within the circle are equal.

From the point P without the circle ABC let two lines PB, PD be drawn making equal angles with PO, the line joining P and the centre; AB shall be equal to CD.

E

B

P

Let fall the perpendiculars OE, OF; then since the angle at E is equal to the angle at F, and EPO=FPO, and the side PO, opposite to one of the equal angles in each is common, .. OE OF, and consequently (Eucl. iii. 14.) ABCD.

=

(3.) Of all straight lines which can be drawn from two given points to meet on the convex circumference of a given circle; the sum of those two will be the least, which make equal angles with the tangent at the point of concourse.

Let A and B be two given points, CE a tangent to the circle at C, where the lines AC, BC make equal angles with it; and let lines AD, BD be drawn from A

and B to any other point D on the

E

B

convex circumference; AC and CB together are less than AD, DB together.

D

GEOMETRICAL PROBLEMS.

[Sect. 2. Let AD meet the tangent in E. Join EB, then (i. 6.) AC and CB together are less than AE and EB; but AE, EB are less than AD, DB (Eucl. i. 19.), .. a fortiori AC, CB are less than AD, DB. And the same may be proved of lines drawn to every other point in the convex circumference.

(4.) If a circle be described on the radius of another circle; any straight line drawn from the point where they meet, to the outer circumference, is bisected by the

interior one.

Let ADB be a circle described on the radius AB of the circle ACE. Draw any line AC meeting the circle ABD in D; AD is equal to DC. Join DB. Then the angle ADB being in a semicircle is a right angle;

and therefore BD being drawn from the centre B of the circle ACE bisects AC (Eucl. iii. 3.).

(5.) If two circles cut each other, and from either point of intersection diameters be drawn; the extremities of these diameters and the other point of intersection shall be in the same straight line.

Let the two circles ABC, ABD cut each other in A and B, draw the diameters AC, AD, and join BC, BD; CB and BD are in the same straight line.

Join AB; the angles ABC, ABD being angles in semicircles are right angles, and therefore (Eucl. i. 13.) CB and BD are in the same straight line.

B

(6.) If two circles cut each other, the straight line joining their two points of intersection, is bisected at right angles by the straight line joining their centres.

Join BD, DA, AC, CB.

Since AD=DB and DC is common to the triangles ADC, BDC and the base AC= CB, ... (Eucl. i. 8.) the angle ADE = BDE. Hence the two sides AD, DE are equal to the two BD, DE and the included angles are equal, .. (Eucl. i. 4.) AE = EB, and the angle DEA=DEB, and being adjacent, they are right angles, i. e. DC bisects AB at right angles.

(7.) To draw a straight line which shall touch two given circles.

1. If the circles be equal.

Let A and B be the centres, join AB; and from A

Α

and B draw AC, BD at right angles to it; join CD. Then AC being parallel and equal to DB; CD is parallel to AB, . CABD is a rectangular parallelogram; and the angles at C and D being right angles, CD is a tangent to both circles (Eucl. iii. 16. Cor.).