5-15-1

Special Segments in TrianglesSpecial Segments in Triangles

I. Triangles have four types of I. Triangles have four types of special segments:special segments:

A. Perpendicular bisectorA. Perpendicular bisector

Any point on the perpendicular bisector of Any point on the perpendicular bisector of a segment is equidistant from the a segment is equidistant from the endpoints of the segment endpoints of the segment

every triangle has 3 perpendicular bisectors

Examples of perpendicular bisectors

In a right triangle, the perpendicularbisectors meet on the triangle.

In an acute triangle, the perpendicular bisectors meet inside the triangle.

In an obtuse triangle, theperpendicular bisectors meet outside the triangle.

CircumcenterCircumcenter

The point where all the perp. Bisectors The point where all the perp. Bisectors meet in a trianglemeet in a triangle

B. MedianB. Median

A median of a triangle is asegment whose endpoints area vertex of the triangle and themidpoint of the opposite side.

D

PROPERTIES OF MEDIANSPROPERTIES OF MEDIANS

• every triangle has 3 medians

• the medians always meet inside the triangle

CentroidCentroid

Point where all the medians meetPoint where all the medians meet

C. AltitudeC. Altitude

The three altitudes intersect at G,a point inside the triangle.

A

B C

In right triangle ABC, segment AB and segment BC are two of the altitudes of the triangle

OrthocenterOrthocenter

Point where all the altitudes meetPoint where all the altitudes meet

D. Angle bisectorD. Angle bisector

-there are 3 angle bisectors in every triangle.

-any point on an angle bisector of an angle is equidistant (same distance) from the sides of the angle.

-any point on or in the interior of an angle and equidistant from the sides of the angle lies on the angle bisector.

-if a triangle is isosceles, the bisector of the vertex angle is also a median and an altitude.

-The angle bisectors always meet inside the triangle.

IncenterIncenter

Point where all the angle bisectors meetPoint where all the angle bisectors meet

1. Given: < F = 80, < E = 301. Given: < F = 80, < E = 30DG bisects < EDFDG bisects < EDF

Prove: < DGE = 115Prove: < DGE = 115F

D

G

E

2.2. Find x and < 2.Find x and < 2.MS is an altitude of MS is an altitude of MNQ MNQ

< 1 = 3x + 11< 1 = 3x + 11< 2 = 7x + 9< 2 = 7x + 9

M

R

QS N

1

2

3.3. MS is a median of MS is a median of MNQ. MNQ.QS = 3a -14 SN = 2 a + 1QS = 3a -14 SN = 2 a + 1<MSQ = 7a + 1 Find a.<MSQ = 7a + 1 Find a.

Is MS an altitude?Is MS an altitude?M

R

QS N

1

2

4. Always, sometimes, never4. Always, sometimes, never

3 medians intersect in the interior3 medians intersect in the interior 3 altitudes intersect at a vertex3 altitudes intersect at a vertex 3 angle bisectors intersect on the exterior3 angle bisectors intersect on the exterior 3 perpendicular bisectors intersect on the exterior3 perpendicular bisectors intersect on the exterior

5. Triangle ABC has vertices A(-3, 10) B(9, 2) and C(9, 15)

a. Determine the coordinates of P on segment AB so that segment CP is

a median of the triangleb. Determine if segment CP is also an altitude of the

triangle

C

A B

a. P is the midpoint of segment AB (-3 + 9)/2 (10 + 2)/2 The midpoint is P(3, 6)

b. The slope of segment CP is (6 – 15)/(3 – 9) = 3/2 The slope of segment AB is (2 – 10)/(9 - - 3) = -2/3 Since the slopes are opposite reciprocals, CP is perpendicular to AB. Therefore, segment CP is an altitude B

6. Find the indicated information:

a. P

Q RS

Find PQ if segment PS isa median of the triangleQS = x + 5 SR = 3x – 17And PQ = 2x -8

b. A

B C D

Find BD if AC is an altitude and m<ACD = 3x + 30, BC = x + 4 and CD = 2x + 8

c.

X Y Z

W

Find m<XWZ if m<XWY = 2x – 4 and m<XWZ = 5x – 12. Segment WY is an anglebisector.

Solutions to example 6:

a. Since segment PS is a median, QS = SR. So, x + 5 = 3x – 17. Then 5 = 2x – 17 22 = 2x 11 = x Finally PQ = 2x – 8 2(11) – 8 PQ = 14

b. If BC is an altitude, then <ACB is a right angle. Then, 3x + 30 = 90. 3x = 60 x = 20 Finally BD = BC + CD x + 4 + 2x + 8 3x + 12 3(20) +12 BD = 72

c. Since segment WY is an angle bisector, it divides the angle in half. So m<XWZ = 2m<XWY. Therefore 5x – 12 = 2(2x - 4) 5x – 12 = 4x - 8 x – 12 = -8 x = 4 So m<XWZ = 5x – 12 5(4) – 12 m<XWZ = 8

Example 7: Write at least one conclusion that can be made from each of the following statements.

S

N

EMR

La. Segment SM is an altitude to segment REb. SN = NEc. M is equidistant from R and E, and <RMS is a right angled. m<ERN = m<SRNe. segment EL is perpendicular to segment SR

Sample answers:

a. Segment SM is perpendicular to segment RE <SMR and <SME are right anglesb. segment RN is a median and N is the

midpointc. Segment SM is a perpendicular bisector segment SM is a median segment SM is an altituded. Segment RN is an angle bisectore. Segment EL is an altitude and <ELR and

<ELS are right angles

1. The segment that bisects an angle of 1. The segment that bisects an angle of the the and has one endpoint at a vertex of and has one endpoint at a vertex of

the the and the other endpoint at another and the other endpoint at another point on the point on the is called the _________. is called the _________.

2.  A _________ is a line or segment that 2.  A _________ is a line or segment that passes through the midpoint of a side of a passes through the midpoint of a side of a

and is perpendicular to that side. and is perpendicular to that side.

3. Triangle XYZ  has vertices X(-1, 1), 3. Triangle XYZ  has vertices X(-1, 1),

Y(3, 9), and Z(6, -2). Determine theY(3, 9), and Z(6, -2). Determine thecoordinates of point W on so that is a coordinates of point W on so that is a median of the triangle. median of the triangle.

4. Triangle CPR has vertices C(15, 1),4. Triangle CPR has vertices C(15, 1),

P(9, 11), and R(2, 1). Determine the P(9, 11), and R(2, 1). Determine the coordinates of point A on so that is a coordinates of point A on so that is a

median of triangle CPR. median of triangle CPR.

5-2 Inequalities for the sides 5-2 Inequalities for the sides and angles of a triangleand angles of a triangle

I. TheoremI. TheoremThe angle opposite the larger side is The angle opposite the larger side is

always bigger than the angle opposite the always bigger than the angle opposite the shorter side of any triangleshorter side of any triangle

II. Theorem II. Theorem

If one angle is larger than another angle, If one angle is larger than another angle, then the side opposite it is also larger than then the side opposite it is also larger than the side opposite the smaller angle.the side opposite the smaller angle.

III. TheoremIII. Theorem

The perpendicular segment from a point to The perpendicular segment from a point to a line is the shortest segment from the a line is the shortest segment from the point to the line.point to the line.

CorollaryCorollary The perp. Segment from a point The perp. Segment from a point to a plane is the shortest segment also.to a plane is the shortest segment also.

IV.ExamplesIV.Examples

1. Refer to the figure in example 1 of 1. Refer to the figure in example 1 of book.book.

Given: angle A is greater than angle DGiven: angle A is greater than angle DProve: BD is greater than ABProve: BD is greater than AB

2. Draw triangle JKL with J(-4,2) K (4,3) L 2. Draw triangle JKL with J(-4,2) K (4,3) L (1,-3). List the angles from greatest to (1,-3). List the angles from greatest to least measure.least measure.

5-3 Indirect Proof5-3 Indirect Proof

I.I. Indirect Proof StepsIndirect Proof Steps Assume that the conclusion is false.Assume that the conclusion is false. Show that this leads to a Show that this leads to a

contradiction of a known property, rule, contradiction of a known property, rule, etc…etc…

Point this out and since it is false, it Point this out and since it is false, it follows that the conclusion must be true.follows that the conclusion must be true.

You are proving that the contrapositive You are proving that the contrapositive of the conditional is true, therefore of the conditional is true, therefore the original conditional must be truethe original conditional must be true

II. A. II. A. Exterior Angle Inequality Exterior Angle Inequality TheoremTheorem

If an angle is an exterior angle of a If an angle is an exterior angle of a triangle, then its measure is greater triangle, then its measure is greater than the measure of either of its than the measure of either of its corresponding remote interior angles.corresponding remote interior angles.

Recall that the sum of the two remote interior angles is Recall that the sum of the two remote interior angles is equal to the exterior. Since neither of the interior equal to the exterior. Since neither of the interior angles are zero, the exterior angle will always be angles are zero, the exterior angle will always be bigger than either of them.bigger than either of them.

                                   

B. Definition of an inequalityB. Definition of an inequality

For any real numbers a and b, a > b if For any real numbers a and b, a > b if and only if there is a positive number c and only if there is a positive number c such that a = b + c.such that a = b + c.

Properties of Inequalities for Properties of Inequalities for Real Numbers p. 254 in bookReal Numbers p. 254 in book

Comparison Property:Comparison Property:

one of three things has to be trueone of three things has to be true

a<b a>b a=ba<b a>b a=b

Transitive Property:Transitive Property:

If a<b and b<c, then a<cIf a<b and b<c, then a<c

If a>b and b>c, then a>cIf a>b and b>c, then a>c

Addition and Subtraction Property:Addition and Subtraction Property:

if a>b then a+c>b+c and a-c>b-cif a>b then a+c>b+c and a-c>b-c

if a<b then a+c<b+c and a-c<b-cif a<b then a+c<b+c and a-c<b-c

Multiplication and Division PropertyMultiplication and Division Property if c>o (positive)and a>b then ac>bc and a/c>b/cif c>o (positive)and a>b then ac>bc and a/c>b/c

if c>o (positive)and a<b then ac<bc and a/c<b/cif c>o (positive)and a<b then ac<bc and a/c<b/c

if c<o (negative) and a>b then ac<bc and a/c<b/cif c<o (negative) and a>b then ac<bc and a/c<b/c

if c<o (negative) and a<b then ac>bc and a/c>b/cif c<o (negative) and a<b then ac>bc and a/c>b/c

REMEMBER:if you multiply or divide an inequality by a REMEMBER:if you multiply or divide an inequality by a negative number it switches the direction of the inequalitynegative number it switches the direction of the inequality

III. Examples III. Examples

1.  Which assumption would you make to 1.  Which assumption would you make to start an indirect proof of the statement start an indirect proof of the statement “two acute angles are congruent”.“two acute angles are congruent”.

2.   Which assumption would you make to 2.   Which assumption would you make to start an indirect proof of the following start an indirect proof of the following statements?statements?

Bob took the dog for a walkBob took the dog for a walkEF is not a perpendicular bisectorEF is not a perpendicular bisector3x = 4 y + 13x = 4 y + 1< 1 is less than or equal to < 2< 1 is less than or equal to < 2

3. Name the property that justifies if 3. Name the property that justifies if aa < < bb, , then then aa + + cc < < bb + + cc. .

4. Name the property that justifies that if 4. Name the property that justifies that if aa is less than is less than bb, then , then aa cannot be greater cannot be greater than than bb. .

5. Given: 2 y + 8 = 165. Given: 2 y + 8 = 16

Prove: y Prove: y ≠ 5≠ 5

6. Given: JKL with side lengths 3, 4, 56. Given: JKL with side lengths 3, 4, 5

Prove: < K Prove: < K < < L< < L

J

K L

Refer to page 255 in your Refer to page 255 in your textbook and try 5 –13textbook and try 5 –13

(hint for 8 – 10: use the (hint for 8 – 10: use the exterior angle inequality exterior angle inequality

theorem-the exterior angle is theorem-the exterior angle is greater than the remote greater than the remote

interior angles)interior angles)

Answers to p 255-256Answers to p 255-2565 -135 -13

5. Assume lines l and m do not intersect at x5. Assume lines l and m do not intersect at x

6. Assume: If the alt int <‘s formed by two parallel lines6. Assume: If the alt int <‘s formed by two parallel lines

and a transversal are congruent, then the lines areand a transversal are congruent, then the lines are

not parallel.not parallel.

7. Assume Sabrina did not eat the leftover pizza7. Assume Sabrina did not eat the leftover pizza

8. <3, <7, <5, <6 9. <4 and <88. <3, <7, <5, <6 9. <4 and <8

10. <8 > <7 because <7 is part of a remote interior angle for 10. <8 > <7 because <7 is part of a remote interior angle for the exterior <8 the exterior <8

11. Division 12. Addition 11. Division 12. Addition 13. Transitive13. Transitive

5-4 The Triangle Inequality5-4 The Triangle Inequality

I. Theorem I. Theorem The sum of the lengths of any two sides of The sum of the lengths of any two sides of

a triangle is greater than the length of the a triangle is greater than the length of the third side.third side.

II. ExamplesII. Examples

1. If Mrs. Ewing gave Elizabeth four 1. If Mrs. Ewing gave Elizabeth four pieces of tubing measuring 6 m, 7 m, 9 m, pieces of tubing measuring 6 m, 7 m, 9 m, and 16 m, how many different triangles and 16 m, how many different triangles could she make?could she make?

2. What are the possible lengths for the 2. What are the possible lengths for the third side of a triangle with two sides of 8 third side of a triangle with two sides of 8 and 13?and 13?

3. How many triangles can be made from 3. How many triangles can be made from a rope 10 ft long?a rope 10 ft long?

5-5 Inequalities with two 5-5 Inequalities with two trianglestriangles

I. TheoremI. TheoremSAS InequalitySAS Inequality

II.II. Theorem Theorem

SSS InequalitySSS Inequality

5-2 Right Triangles5-2 Right Triangles

I. What are the postulates for proving 2 I. What are the postulates for proving 2 right triangles are congruent?right triangles are congruent?

SAS aka LLSAS aka LL

LL means Leg Leg

ASA, AAS aka LA ASA, AAS aka LA

LA means Leg Angle

AAS aka HA AAS aka HA

HA means Hypotenuse Angle

Finally, a new one!Finally, a new one!

HL (only for right HL (only for right triangles/in non-right triangles/in non-right triangles SSA)triangles SSA)

HL means Hypotenuse Leg

1. In the figure,      is the angle bisector < BAC. Are the triangles congruent?

 

  

             

Tell which of the righttriangle methods will prove the trianglescongruent.

2. Find x so that the right triangles 2. Find x so that the right triangles are congruent.are congruent.

3. What do you need to prove by 3. What do you need to prove by HA?HA?

4. Name the theorem used to 4. Name the theorem used to prove the triangles are congruent.prove the triangles are congruent.

5. Which theorem or postulate states 5. Which theorem or postulate states that if the hypotenuse and an acute that if the hypotenuse and an acute angle of one right triangle are angle of one right triangle are congruent to the hypotenuse and congruent to the hypotenuse and corresponding acute angle of another corresponding acute angle of another right triangle, then the triangles are right triangle, then the triangles are congruent?congruent?

Refer to page 248 in your textbook and try problems 6-11

Answers to p 248: 6-11

6. Need to know that <B and <D are right angles You already know AC = AC

7. Need to know that ST = TU

8.Need to know that LN = QR or NM = RP

9. 2x + 10 = 5x – 8 so 10 = 3x –8 then 18 = 3x and 6 = x

10. x + 7 = 3x – 5 so 7 = 2x – 5 then 12 = 2x and 6 = x

11. 4x – 26 = 10 then 4x = 36 and x = 9