In AC make AE=AD, and join DE.

On DE, on the side remote from A, describe an equilat. ▲ DFE.

Join AF. Then AF will bisect BAC.

For in As AFD, AFE,

· AD=AE, and AF is common, and FD=FE,

.. L DAF LEAF,

==

that is, ▲ BAC is bisected by AF.

I. 1.

I. c.

Q. E. F.

Ex. 1. Shew that we can prove this Proposition by means of Prop. IV. and PROP. A., without applying Prop. C.

Ex. 2. If the equilateral triangle, employed in the construction, be described with its vertex towards the given angle; shew that there is one case in which the construction will fail, and two in which it will hold good.

NOTE.--The line dividing an angle into two equal parts is called the BISECTOR of the angle.

Let AB be the given st. line.

It is required to bisect AB.

On AB describe an equilat. ▲ ACB.

Bisect ACB by the st. line CD meeting AB in D;

then AB shall be bisected in D.

For in As ACD, BCD,

· AC=BC, and CD is common, and ▲ ACD= ▲ BCD,

:. AD=BD ;

.. AB is bisected in D.

I. 1.

I. 9

I. 4.

Q. E. F.

Ex. 1. The straight line, drawn to bisect the vertical angle of an isosceles triangle, also bisects the base.

Ex. 2. The straight line, drawn from the vertex of an isosceles triangle to bisect the base, also bisects the vertical angle.

Ex. 3. Produce a given finite straight line to a point, such that the part produced may be one-third of the line, which is made up of the whole and the part produced.

PROPOSITION XI. 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 st. line, and C a given pt. in it.
It is required to draw from C a st. line 1 to AB.

Take any pt. D in AC, and in CB make CE=CD.
On DE describe an equilat. ▲ DFE.

Join FC. FC shall be 1 to AB.

For in As DCF, ECF,

I. 1.

·· DC=CE, and CF is common, and FD=FE,

.. 4 DCFL ECF;

and.. FC is 1 to AB.

I. c.

Def. 9.

Q. E. F.

COR. To draw a straight line at right angles to a given straight line AC from one extremity, C, take any point D in AC, produce AC to E, making CE=CD, and proceed as in the proposition.

Ex. 1. Shew that in the diagram of Prop. ix. AF and ED intersect each other at right angles, and that ED is bisected by AF.

Ex. 2. If O be the point in which two lines, bisecting AB and AC, two sides of an equilateral triangle, at right angles, meet; shew that OA, OB, OC are all equal.

Ex. 3. Shew that Prop. XI. is a particular case of Prop. IX.

PROPOSITION XII. PROBLEM.

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

A

В

Let AB be the given st. line of unlimited length; the given pt. without it.

It is required to draw from C a st. line 1 to AB.

Take any pt. D on the other side of AB.

With centre C and distance CD describe a O cutting AB

in E and F.

Bisect EF in 0, and join CE, CO, CF.

Then CO shall be to AB.

I. 10.

For in AS COE, COF,

· EO=FO, and CO is common, and CE=CF,

.. 4 COEL COF;

.. CO is to AB.

I. c.

Def. 9.

Q. E. F.

Ex. 1. If the straight line were not of unlimited length, how might the construction fail?

Ex. 2. If in a triangle the perpendicular from the vertex on the base bisect the base, the triangle is isosceles.

Ex. 3. The lines drawn from the angular points of an equilateral triangle to the middle points of the opposite sides are equal.

Miscellaneous Exercises on Props. I. to XII.

1. Draw a figure for Prop. II. for the case when the given point A is

(a) below the line BC and to the right of it.

(B) below the line BC and to the left of it.

2. Divide a given angle into four equal parts.

3. The angles B, C, at the base of an isosceles triangle, are bisected by the straight lines BD, CD, meeting in D; shew that BDC is an isosceles triangle.

4. D, E, F are points taken in the sides BC, CA, AB, of an equilateral triangle, so that BD-CE=AF. Shew that the triangle DEF is equilateral.

5. In a given straight line find a point equidistant from two given points; 1st, on the same side of it; 2d, on opposite sides of it.

6. ABC is a triangle having the angle ABC acute. In BA, or BA produced, find a point D such that BD=CD.

7. The equal sides AB, AC, of an isosceles triangle ABC are produced to points F and G, so that AF-AG. BG and CF are joined, and H is the point of their intersection. Prove that BH=CH, and also that the angle at A is bisected by AH.

8. BAC, BDC are isosceles triangles, standing on opposite sides of the same base BC. Prove that the straight line from A to D bisects BC at right angles.

9. In how many directions may the line AE be drawn in Prop. III.?

10. The two sides of a triangle being produced, if the angles on the other side of the base be equal, shew that the triangle is isosceles.

11. ABC, ABD are two triangles on the same base AB and on the same side of it, the vertex of each triangle being outside the other. If AC=AD, shew that BC cannot =BD.

12. From Cany point in a straight line AB, CD is drawn at right angles to AB, meeting a circle described with centre A and distance AB in D; and from AD, AE is cut off = AC: shew that AEB is a right angle.