MISCELLANEOUS EXAMPLES ON BOOK I.

1. Prove that the angle between the bisectors of two consecutive angles of a quadrilateral is equal to half the sum of its other two angles.

2. Prove that the sum of the distances of any point taken within an equilateral triangle from the three sides is the same whereon the point may be. Investigate the case in which the point is without the triangle.

3. AD and BC are two parallel straight lines cut obliquely by AB and perpendicularly by AC, and the straight line BED is drawn between them, cutting AC in E, so that ED is equal to twice AB. Prove that the angle DBC is one third of ABC.

4. On the side AB of the triangle ABC take AD equal to AC, produce BA and take AE equal to AC, then join the angle C to the points D and E. Prove that the angle at E is half BAC and that DCE is a right angle.

5. Prove that the point of intersection of the straight lines joining the middle points of opposite sides of any quadrilateral is the middle of the straight line joining the middle points of the diagonals.

6. ABC is a triangle right-angled at A, and ABDM and ACEN are squares upon AB and AC respectively. From the points D and E perpendiculars DF and EG are drawn to the hypothenuse BC produced. Prove that the hypothenuse BC is equal to the sum of the perpendiculars DF and EG.

7. In every trapezoid the middle points of the two non-parallel sides and of the two diagonals are on one straight line which, parallel to the other two sides and the distance between the two extreme points, is equal to half the sum of these sides and the distance between the intermediate points to half their difference.

8. Prove that if a ball lying on a rectangular billiard

table be set in motion in a direction parallel to one of the diagonals of the table, it will, after rebounding from each cushion in succession, return to the point of departure, and that the path described by it will be equal to the sum of the diagonals of the table.

9. If on the sides of a square ABCD any four points E, F, G, and H be taken, one on each side, so that AE BF = CG = DH, prove that EFGH will also be a

square.

10. If through any point on the base of an isosceles triangle parallels be drawn to the other two sides, prove that the parallelogram so formed will have the same perimeter wherever the point may be situated.

BOOK II.

ON THE CIRCLE.

SECTION I.-MISCELLANEOUS PROPOSITIONS.

DEFINITIONS.

35.-The locus of a point which is always in one plane and at the same given distance from a given point in that plane is a circle; the given point is called the centre, and the given distance is called the radius of the circle.

36.-The portion of the circle between any two points upon it is called an arc of the circle, and the straight line joining the two points is called the chord of the arc.

N.B. The line forming the locus is sometimes called the circumference of the circle and sometimes the circle.

PROPOSITION 1.

A circle is a finite closed figure, such that every point in the plane of the circle whose distance from the centre is less than the radius of the circle is situated within the circle; and every point in the plane of the circle whose distance from the centre is greater than the radius of the circle is situated without the circle.

Let A be the centre of any given circle.

Draw the indefinite straight line AB terminated at A, but unlimited towards B.

Because AB is a straight line of indefinite length, therefore it is always possible to find a point P upon AB, such that AP is equal to the radius of the circle.

Fig. 1.

Also every point on AB between P and B is at a distance from A greater than the radius of the circle; and every point on AB between A and P is at a distance from A less than the radius of the

circle.

Therefore every in

⚫ definite straight line

Эт

B

drawn from A meets the circle in one point, and one point only.

Therefore the circle is a finite closed figure, such that every point in the plane of the circle whose distance from A is less than the radius of the circle is situated within the circle; and every point in the plane of the circle whose distance from A is greater than the radius of the circle is situated without the circle.

PROPOSITION 2.

Two circles which have the same centre and one point in common must coincide in every point.

Let, if possible, AB and AC be non-coincident arcs of two circles having the same centre D and the point A common to both.

Draw any straight line through D, meet ing these arcs in B and C, and join DA.

Because D is the centre of the circle of which AB is an arc, therefore DA is equal to DB.

Similarly, DA is equal to DC,

therefore DB is equal to DC,
which is impossible.

A

Fig. 2.

B

C

D

PROPOSITION 3.

If the perpendicular drawn to any indefinite straight line from the centre of a given circle be less than the radius of the circle, this straight line shall meet the circle in two points, and two points only, and these points shall be equidistant from the foot of the aforesaid perpendicular.

Let C be the centre of the circle GEF, and let AB be an indefinite straight line such that CD the perpendicular

drawn from C to AB is less than the radius of GEF, then AB shall meet GEF in two points, and two points only, equidistant from D.

Because DB is an indefinite straight line, therefore it is always possible to find a point H in DB such that DH is equal to the radius of the circle GEF.

Join CH.

Because CDH is a right angle, therefore CH is greater than DH (Bk. I. Prop. 15),

therefore CH is greater than the radius of the circle GEF, therefore the point H is situated without the circle GEF (Bk. II. Prop. 1).

Again, because CD is less than the radius of the circle GEF, therefore the point D is situated within the circle GEF.

Because the circle GEF is a finite closed figure, therefore the straight line DH which joins the point D within this