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119. Axiom IX. If a, b, c, d are line-segments (or angles) such that a> b and c d or such that a>b and c>d, then a+c>b+d. Also if a>b and c≤d, then a-c>b-d, provided a >c and b>d. (See §§ 10, 39.)

120. THEOREM. The sum of the segments drawn from a point within a triangle to the extremities of one side is less than the sum of the other two sides.

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.. AO+OE + OB < AC + CE + OE+ EB. (Ax. IX)

Subtracting OE from both members,

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1. Show that any side of a triangle is greater than the difference between the other two sides.

2. Show that the sum of the distances from any point within a triangle to the vertices is greater than one-half the sum of the sides of the triangle.

3. Show that the segment joining the vertex of an isosceles triangle to any point in the base is less than either of the equal sides.

4. Show that any altitude of an equilateral triangle bisects the vertex angle from which it is drawn and also bisects the base. 5. State and prove the converse of the theorem in Ex. 4.

6. Construct angles of 60°, 120°, and 30°.

7. Construct angles of 45° and 135°.

SUGGESTION. Bisect a right angle and extend one side.

122. PROBLEM. Given two sides of a triangle and an angle opposite one of them, to construct the triangle.

a

B

K

SOLUTION. Let A and the segments a and b be the given parts. On one side of A lay off AB= = b, and let the other side be extended to some point K.

With B as a center and a radius equal to a, construct arcs of circles as shown in the figure.

The following cases are possible:

(1) If a equals the perpendicular distance from B to AK, the arc will meet AK in but one point, and a right triangle is the solution.

(2) If a <b and greater than the perpendicular, the arc cuts AK in two points, and there are then two triangles containing the given parts, as shown by the dotted lines in the figure.

(3) If a > b, the arc will cut AK only once on the right of 4, and hence only one triangle will be found. Repeat this construction, making a separate figure for each case.

Make the construction when A is a right angle. Are all three cases possible then? Make the construction when A is obtuse. What cases are possible then?

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1. A carpenter bisects an angle A as follows: Lay off AB AC. Place a steel square so that

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Draw the line AD. Is this method
correct? Give proof. Would this
method be correct if the square
were not right-angled at D?

2. In the triangle ABC, AC = BC. The points B D, E, F are so placed that AD = BD and AF = BE. Compare DE and DF. Prove your conclusion.

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3. If in the figure 2 21 15°, and 24 = find each angle of the triangle.

120,

A A A A

B

4. If in A ABC, AC = BC, and if AC is extended to D so that

AC = CD, prove that DB is perpendicular to AB.

5. In ▲ ABC, CA =

CB, AD = BE. Prove ▲ ADB ≈ ▲ ABE.

6. In the triangle KLN, NM is perpendicular to KL, and KM = MN ML. Prove that KLN is an isosceles

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right triangle.

7. If in the isosceles ▲ ABC a point D lies in the base, and 21 = 42, determine whether there is any position for D such that DE = DF.

8. If the bisectors AD and BE of the base angles of an isosceles triangle ABC meet in 0, what pairs of equal angles are formed? What pairs of equal segments? Of congruent triangles?

9. If the middle points of the sides of an equilateral triangle are connected as shown in the figure, compare the resulting four triangles.

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10. Triangle ABC is equilateral. ADBE = CF. Compare the triangles DBE, ECF, FAD.

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11. Two railway tracks cross as indicated in the figure. What angles are equal and what pairs of angles are supplementary? State a theorem involved in each case.

12. In a AABC does the bisector of A also bisect the side BC (1) if AC = BC but AC <AB, (2) if AC = AB?

E

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13. If in the triangle ABC, AB = AC and if E is any point on AC, find D on AB so that

AECBA CDB.

14. If in the figure AB = AC, find 21 if

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follows: Place two squares against the timbers, as shown in the figure, so that AO BO. Show that AB is the required line.

DETERMINATION OF LOCI.

124. THEOREM. Every point on the bisector of an angle is equidistant from the sides of the angle.

Given P, any point on the bisector of the

angle A, and PC and PB perpendicular to the sides.

To prove that PC = PB.

Proof: By the hypothesis

21 = 22.

Show that 3 = 24 and complete the proof.

3

P

125. THEOREM. If a point is equally distant from the sides of an angle, it lies on the bisector of the angle.

Given an angle A, any point P and the perpendiculars PB and PC equal.

To prove that PA bisects the angle A. Proof: Give the argument in full to show that ΔΑΒΡΩΔΑPC and thus show that 1= 22.

Hence AP is the bisector of the angle 4.

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126. The two preceding theorems enable us to assert the following:

(1) Every point in the bisector of an angle is equidistant from its sides.

(2) Every point equidistant from the sides of an angle lies in its bisector.

For these reasons the bisector of an angle is called the locus of all points equidistant from its sides.

The word locus means place or position. It gives the location of all points having a given property.

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