a circle, when its distance from the centre is less than the radius of the circle.

It is evident that

(1) If a point in a straight line is within a closed figure the straight line if produced in either sense from the point will meet the boundary of the figure, and thus intersect it in two points at least ;

(2) If a point in the boundary of one closed figure lie within another closed figure, and also a point in the boundary of the latter lie within the former, the two boundaries intersect in two points at least ;

For they cannot lie wholly outside each of the other and if one were wholly inside the other, no point in the boundary of the second would lie within the first. Hence they must lie partly inside and partly outside each of the other, and their boundaries (circumferences) must meet in two points at least.

By the help of the above, it may be shewn that the straight lines and circles drawn in the Problems of this section intersect; or the conditions that must be satisfied in order that they may intersect may be determined.

PROB. 1. To bisect a given angle.

Let BAC be the given angle:

it is required to bisect it.

B

D

With centre A, and with any radius, draw a circle cutting AB at D, and AC at E. Post. 3.

With centres D and E, and with any radius greater than half the straight line DE, draw circles,

Post. 3.

let F be a point of intersection of these circles which lies

within the angle BAC;

join AF :

then shall AF bisect the angle BAC.

Post. I.

Join DF, FF.

Then in the triangles DAF, EAF,

the side AD is equal to the side AE, the side AF is common to both,

and the side DF is equal to the side EF,

therefore the angle DAF is equal to the angle EAF, therefore AF bisects the angle BAC.

Constr.

Constr.

I. 18.

Q.E.F.

Ex. 74. Divide a given angle into four equal parts. Ex. 75. Prove that FA produced will bisect the major conjugate angle BAC.

PROB. 2. To draw a perpendicular to a given straight line from a given point in it.

Let BAC be the given straight line, A the given point in it :

it is required to draw from A a perpendicular to BAC.

A

With centre A, and with any radius, draw a circle cutting

AB at D, and AC at E,

Post. 3.

With centres D and E, and with any radius greater than AD or

AE, draw circles,

Post. 3.

let F be a point of intersection of these circles;

and the side DF is equal to the side EF,

therefore the angle DAF is equal to the angle EAF, therefore AF is perpendicular to BAC.

Constr.

Constr.

I. 18.

Defs. 10, 11.

Q.E.F.

Ex. 76. Shew that Prob. 2 is a particular case of Prob. 1.

PROB. 3. To draw a perpendicular to a given straight line from a given point outside it.

Let AB be the given straight line, C the given point outside it:

it is required to draw a perpendicular to AB from C.

H

Take any point D on the side of AB remote from C,

with centre C and radius CD draw a circle cutting AB produced if necessary at E and F, Post. 3.

with centres E and F and the same radius as before draw circles cutting one another, on the side of AB remote from C, at G,

Post. 3.

join CG cutting AB at H :

then shall CH be perpendicular to AB.

Join CE, CF, EG, FG.

Then in the triangles ECG, FCG,

the side CE is equal to the side CF,

the side EG is equal to the side FG,

and the side CG is common to both,

therefore the angle ECG is equal to the angle FCG.

Again, in the triangles ECH, FCH,

the side EC is equal to the side FC,

the side CH is common to both,

Post. I.

Constr.

Constr.

I. 18.

Constr.

and the angle ECH is equal to the angle FCH,

therefore the angle CHE is equal to the angle CHF,

I. 5.

therefore CH is perpendicular to AB.

Def. 10.

Q.E.F.

Ex. 77. Why is the point D taken on the side of AB remote

from C?

Ex. 78. Shew, by Theor. 15, Cor., that the circle whose centre is C cannot meet AB in more than two points.

PROB. 4. To bisect a given finite straight line.

Let AB be the given finite straight line

it is required to bisect it.

With centres A and B and radii equal to AB draw circles cutting one another at C and D,

Post. 3.

and the side CD is common to both,

therefore the angle ACD is equal to the angle BCD.

Again, in the triangles ACE, BCE,

the side AC is equal to the side BC,

the side CE is common to both,

and the angle ACE is equal to the angle BCE, therefore the side AE is equal to the side BE, therefore AB is bisected at E.

I. 18.

Constr.

I. 5.

Q.E.F.