PROPOSITION 9. PROBLEM.

To bisect a given angle, that is, to divide it into two equal

parts.

A

E

B

Let BAC be the given angle:
it is required to bisect it.

Construction. In AB take any point D;
and from AC cut off AE equal to AD.
Join DE;

and on DE, on the side
lateral triangle DEF.

I. 3.

remote from A, describe an equi

Join AF.

Then shall the straight line AF bisect the angle BAC.

Proof. For in the two triangles DAF, EAF,

Because

DA is equal to EA,

and AƑ is common to both;

I. 1.

Constr.

and the third side DF is equal to the third side

EF;

Def. 19.
I. 8.

therefore the angle DAF is equal to the angle EAF. Therefore the given angle BAC is bisected by the straight line AF.

Q.E.F.

EXERCISES.

1. If in the above figure the equilateral triangle DFE were described on the same side of DE as A, what different cases would arise? And under what circumstances would the construction fail?

2. In the same figure, shew that AF also bisects the angle DFE.

3. Divide an angle into four equal parts.

PROPOSITION 10. PROBLEM.

To bisect a given finite straight line, that is, to divide it into two equal parts.

Let AB be the given straight line:

it is required to divide it into two equal parts.

Constr. On AB describe an equilateral triangle ABC, and bisect the angle ACB by the straight line CD, meeting AB at D.

I. 1.

I. 9.

Then shall AB be bisected at the point D.

For in the triangles ACD, BCD,

Def. 19.

and CD is common to both;

Proof.

Because

AC is equal to BC,

also the contained angle ACD is equal to the contained angle BCD;

Constr.

Therefore the triangles are equal in all respects: so that the base AD is equal to the base BD. I. 4. Therefore the straight line AB is bisected at the point D.

Q. E. F.

EXERCISES.

1. Shew that the straight line which bisects the vertical angle of an isosceles triangle, also bisects the base.

2. On a given base describe an isosceles triangle such that the sum of its equal sides may be equal to a given straight line.

PROPOSITION 11. PROBLEM.

To draw a straight line at right angles to a given straight line, from a given point in the same.

Let AB be the given straight line, and C the given point in it.

It is required to draw from the point C a straight line at right angles to AB.

Construction. In AC take any point D,

and from CB cut off CE equal to CD.
On DE describe the equilateral triangle DFE.

Join CF.

I. 3.

1. 1.

Then shall the straight line CF be at right angles to AB.

Proof.

Because

For in the triangles DCF, ECF,

DC is equal to EC,

and CF is common to both;

Constr.

and the third side DF is equal to the third side

EF:

Def. 19. Therefore the angle DCF is equal to the angle ECF: 1. 8. and these are adjacent angles.

But when a straight line, standing on another straight line, makes the adjacent angles equal to one another, each of these angles is called a right angle;

Def. 7. therefore each of the angles DCF, ECF is a right angle. Therefore CF is at right angles to AB,

and has been drawn from a point C in it.

Q.E.F.

EXERCISE.

In the figure of the above proposition, shew that any point in FC, or FC produced, is equidistant from D and E.

To draw a straight line perpendicular to a given straight line of unlimited length, from a given point without it.

Let AB be the given straight line, which may be produced in either direction, and let C be the given point without it.

It is required to draw from the point C a straight line perpendicular to AB.

Construction. On the side of AB remote from C take any point D;

and from centre C, with radius CD, describe the circle FDG, meeting AB at F and G.

Bisect FG at H;

and join CH.

Post. 3.

I. 10.

Then shall the straight line CH be perpendicular to AB.

Join CF and CG.

Proof. Then in the triangles FHC, GHC,

Because

FH is equal to GH,

and HC is common to both;

Constr.

and the third side CF is equal to the third side CG, being radii of the circle FDG;

Def. 11. therefore the angle CHF is equal to the angle CHG; 1. 8. and these are adjacent angles.

But when a straight line, standing on another straight line, makes the adjacent angles equal to one another, each of these angles is called a right angle, and the straight line which stands on the other is called a perpendicular to it.

Therefore CH is a perpendicular drawn to the given straight line AB from the given point C without it.

Q. E. F.

NOTE. The given straight line AB must be of unlimited length, that is, it must be capable of production to an indefinite length in either direction, to ensure its being intersected in two points by the circle FDG.

EXERCISES ON PROPOSITIONS 1 TO 12.

1. Shew that the straight line which joins the vertex of an isosceles triangle to the middle point of the base is perpendicular to the base.

2. Shew that the straight lines which join the extremities of the base of an isosceles triangle to the middle points of the opposite sides, are equal to one another.

3. Two given points in the base of an isosceles triangle are equidistant from the extremities of the base: shew that they are also equidistant from the vertex.

4. If the opposite sides of a quadrilateral are equal, shew that the opposite angles are also equal.

5. Any two isosceles triangles XAB, YAB stand on the same base AB: shew that the angle XAY is equal to the angle XBY; and that the angle AXY is equal to the angle BXY.

6. Shew that the opposite angles of a rhombus are bisected by the diagonal which joins them.

7. Shew that the straight lines which bisect the base angles of an isosceles triangle form with the base a triangle which is also isosceles.

8. ABC is an isosceles triangle having AB equal to AC; and the angles at B and C are bisected by straight lines which meet at O: shew that OA bisects the angle BAC.

9. Shew that the triangle formed by joining the middle points of the sides of an equilateral triangle is also equilateral.

10. The equal sides BA, CA of an isosceles triangle BAC are produced beyond the vertex A to the points E and F, so that AE is equal to AF; and FB, EC are joined: shew that FB is equal to EC.

11. Shew that the diagonals of a rhombus bisect one another at right angles.

12. In the equal sides AB, AC of an isosceles triangle ABC_two points X and Y are taken, so that AX is equal to AY; and CX and BY are drawn intersecting in O: shew that

(i) the triangle BOC is isosceles;

(ii) AO bisects the vertical angle BAC;

(iii) AO, if produced, bisects BC at right angles.

13. Describe an isosceles triangle, having given the base and the length of the perpendicular drawn from the vertex to the base.

14. In a given straight line find a point that is equidistant from two given points.

In what case is this impossible?