8. The perimeter of a pentagon is greater than one-half the sum of the diagonals and is less than the sum of the diagonals.

A

A

9. The perimeter of a hexagon is greater than one-half the sum of the diagonals which join alternate angles, and is also greater than two-thirds the sum of the diagonals which join opposite angles.

F

E

(1) AB+BC>AC, etc.

(2) AB+BC+CD>AD, etc.

B

10. All the sides but one of any polygon are together greater than that one.

Use Euc. I. 20.

11. Show that in any quadrilateral the sum of any two sides is greater than the difference of the other two.

Use the method of Ex. 10 or of Ex. 7.

12. If one triangle be wholly enclosed in another, the perimeter of the external triangle shall be greater than that of the internal.

Produce the sides of the internal triangle in order, and use the method of Ex. 10.

13. The sum of the distances of any point from the angular points of a polygon is greater than half the perimeter of the polygon.

14. ABC is a triangle having AB>AC, and D is a point in BC such that AD>AC. AE is bisected in F. BP+PD>BA+AC.

From DA, DE is cut off AC, and
If P be a point in FA, prove that

§ 7. (Bookwork, EUCLID, I. 1-26.)

1. In an isosceles triangle the altitude drawn from the vertex is also the median and the bisector of the vertical angle.

Let ABC be an isosceles triangle having AB-AC, and let AD be BC; it is required to prove that

(1)

BD= CD,

(2) ▲ BAD= < CAD.

AB= AC,

In As ABD, ACD LABD=ACD,

LADBLADC;

... A ABDA ACD;

[Hypothesis. [Euc. I. 5. [Right angles. [Euc. I. 26.

... (1) BD=CD; (2) ▲ BAD= ▲ CAD.

2. In an isosceles triangle the altitudes drawn from the extremities of the base are equal.

DEFINITION. The straight line drawn from a vertex of a triangle, bisecting the angle at that vertex and terminated by the opposite side, is called the Bisector of that angle.

M

3. In an isosceles triangle the bisectors of the angles at the

base are equal.

Use Euc. I. 5 and I. 26 to show

either that Δ ΑΒL=1 ACM,

or ▲ MBC=▲ LCB.

B

4. If in a triangle an altitude be also the bisector of the angle from which it is drawn, the triangle shall be isosceles.

Use Euc. I. 26 to show that

▲ ABDEA ACD.

C

5. P, Q, and R are points in the sides of an equilateral ▲ ABC, such that AP-BQ=CR.

Show that the triangle formed by

the intersection of AQ, BR, CP is equilateral.

Use Euc. I. 4 to show that

B

Q

B

P

R

AABQ=A BCREA CAP;

and Euc. I. 26 to show that

▲ APX A BQY=A CRZ. Hence show that

YZ=ZX=XY.

6. Any point on the bisector of an angle is equally distant from the arms of the angle.

Let BI bisect

Draw IZ, IX

ABC.
AB, BC.

Use Euc. I. 26 to show IX=IZ.

DEFINITION.-The length of the perpendicular drawn from an external point to a straight line is called the Distance of the point from the line.

7. The point in which the bisectors of two angles of a triangle meet is equally distant from the three sides of the triangle.

8. The point in which the bisectors of two exterior angles of a triangle meet is equally distant from the three sides of the triangle.

9. In the ▲ ABC, if D be the mid-point of BC and ADB be obtuse, then AB>AC.

Use Euc. I. 24 to compare

As ADB, ADC.

B

10. The straight lines drawn through the extremities of the base of an isosceles triangle, equally inclined to it and terminated by the sides, are equal.

Use Euc. I. 26.

11. If two triangles have the same base, and if the straight line which joins their vertices bisect both vertical angles, show that both triangles are isosceles.

12. If, in the ▲ ABC, AB>AC and D be the mid-point of BC, then ADB shall be obtuse.

§ 8. (Bookwork, EUCLID, I. 1-26.)

1. Of all straight lines which can be drawn to a given straight line from a given point outside, the perpendicular is the least; and, of the others, those which make equal angles with the perpendicular are equal, and that which makes a greater angle with the perpendicular is greater than that which makes a less angle. (A Standard Theorem.)

Let P be the given point, and AB the given straight line, and let PC be AB, let PD, PE be drawn to AB making equals CPD, CPE, and let PF be < CPF> <CPD; it is required to prove

drawn making

[Euc. I. 17.

[Euc. I. 19.

[Hypothesis. [Right angles.

CPE,

=

▲ PCD PCE;

.. Δ PDC=Δ PEC;

... PD=PE.

[Euc. I. 26.

[Euc. I. 16.

[Euc. I. 17.

[Euc. I. 19.