Every point in AD shall be equally distant from AB,

Let P be any point in AD, and draw PE, PF perpendicular to AB, and to AC.

Then in the triangles PAE, PAF,

PAE=PAF, AEP= AFP, and AP is common.

Therefore

PE=PF

[1. 15.

Also no point not in AD can be equally distant from AB and from AC.

Let G be a point not on AD.

Let fall GE, GH perpendicular to AB, AC.

And let GE meet AD in P; let fall PF perpendicular to AC, and join GF.

Hence AD is the locus of a point equally distant from AB and from AC.

If AD be equally distant from AB and AC, the distance

being measured by perpendiculars let fall from a point in AD, AB and AC are equally remote from AD, the distance being measured along a line perpendicular to AD. For,

E B

GP -D

F -C

joining EF,

each to each, and

EA, AG=FA, AG

EAG=FAG;

therefore EAG, FAG are congruent triangles, and EG = GF. Also the angles EGA, FGA are equal, therefore EF is perpendicular to AD.

The system is thus perfectly symmetrical with regard to AD, for corresponding to every point, line, or angle on the one side of AD, there is a like point, line, or angle on the other.

THEOREM XXI.

The perpendiculars DG, EH, FK drawn through D, E, F,

the middle points of the sides of a triangle ABC, meet in one point,

Let DG and EH intersect in O.

Then since O is a point on DG, it is equally distant from B and C.

[I. 19.

And since O is a point on EH, it is equally distant from C and A.

Therefore O is equally distant from B and A.

Therefore is a point on FK.

That is, FK passes through 0.

[I. 19.

[I. 19.

THEOREM XXII.

The straight lines AD, BE, CF, which bisect the angles of a triangle ABC, meet in one point.

Then since O is a point on AD, it is equally distant from AC and AB.

[I. 20.

And since O is a point on BE, it is equally distant from

BA and BC..

[I. 20.

Therefore is equally distant from CB and CA.

Therefore it is a point on CF.

That is, CF passes through 0.

[1. 20.

THEOREMS.

1. The straight lines which bisect two adjacent supplementary angles are perpendicular to one another.

2. If two straight lines cut one another, and straight lines be drawn bisecting the four angles which they form, these four lines will together constitute two straight lines at right angles to one another.

3. ABC is a triangle, D a point within it, shew that AB+BC+ CA are greater than DA+ DB+DC, and less than twice DA + DB + DC.

4. The two sides of a triangle are together greater than twice the straight line drawn from the vertex to the middle point of the base.

5. If E be a point within the triangle ABC, the angle BEC is greater than the angle BA C. 1. 10. COR. 2.

6. ABC is a triangle, on AB produced if necessary take AC′ = AC, and on AC take AB′ = AB. Join B'C' cutting BC in D. Show that AD bisects BAC.

7. The extremities of the base of an isosceles triangle are equally distant from the opposite sides.

8. The sum of the distances of any point on the base of an isosceles triangle from the sides is constant. What is the

case if the point be taken on the base produced?

9. If the opposite sides of a quadrilateral figure be equal, it is a parallelogram.

10. If the opposite angles of a quadrilateral figure be equal, it is a parallelogram.

11. The diagonals of a parallelogram bisect one another. 12. The diagonals of a rhombus bisect one another at right angles.

13. The converse of the last two propositions.

14. If the diagonals of a parallelogram are equal, it is a rectangle.

15. If the diagonals of a parallelogram are equal and perpendicular, it is a square.

16. Straight lines which bisect the adjacent angles of a parallelogram are perpendicular to one another.

17. ABC is a right-angled triangle, C the right angle. Make ACD=A.

Then BCD = B, and AD=BD = CD.

18. From the angles of a square measure equal distances upon the sides in order. Join the points thus determined. The figure so formed is a square.

19. If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle included by those sides in the one greater than the corresponding angle in the other, the base of the first shall be greater than that of the second.

20. The converse of 19.

21. ABC is a right-angled triangle, A the right angle: draw AD perpendicular to BC: shew that the triangles ABC, ABD are equiangular to one another. Also they have a side common, how is it that they are not congruent?