“There is no success without hardship.” –Sophocles

Concept 1: Definition of a Midsegment

A midsegment of a triangle connects two midpoints on two of the sides of a triangle.

Definitely a midsegment Cannot conclude it is a midsegment

Concept 2: Midsegment Theorem

The segment connecting the midpoints of two sides of a triangle is parallel to the third side and is half as long as that side.

A

B

C D

E

Segment BE is a midsegment of ∆ACD

BE=1

2CD , 𝐵𝐸||𝐶𝐷

Example 1: Find the value of x.

A

B

C D

E

A

B

C D

E

A

B

C D

E

6

x 16

4x

x+4

x-4

Concept 2: Midsegment Theorem

The segment connecting the midpoints of two sides of a triangle is parallel to the third side and is half as long as that side.

A

B

C D

E

Segment BE is a midsegment of ∆ACD

BE=1

2CD , 𝐵𝐸||𝐶𝐷

Segment BF is a midsegment of ∆ACD

BF=1

2AD , 𝐵𝐹||𝐴𝐷

Segment FE is a midsegment of ∆ACD

FE=1

2CA , 𝐹𝐸||𝐶𝐴 F

What lengths are equal in the diagram and which

segments are parallel?

Example 2:

A

B

C D

E

F

BE=CF=FD , 𝐵𝐸||𝐶𝐷 EF=AB=BC , 𝐸𝐹||𝐴𝐶 BF=AE=ED , 𝐵𝐹||𝐴𝐷

Concept 3: Placing of Shapes in a Coordinate Plane

We sometimes need to place an arbitrary shape into a coordinate plane when trying to prove length, midpoint, or slope in a proof.

Conditions for a convenient location:

Has as many 0’s as possible for the coordinates.

Avoids using fractions in a proof.

Uses variables instead of numbers for the coordinates (exception is zero).

Concept 3: Placing of Shapes in a Coordinate Plane

Example of placing a convenient isosceles triangle into a coordinate plane.

(a, 0) (-a, 0)

(0, b)

(a, b) (c, d)

(e, f)

Place these shapes in a convenient location.

1) Right isosceles triangle with leg lengths of b.

2) Scalene triangle with on side length of a

Example 3

(0, b)

(b, 0) (0, 0) (0, 0) (a, 0)

(b, c)

These proofs usually require that length, slope, or

midpoints to be used in the proof. As such they have four distinct steps.

Place the shape in a convenient location.

Find all coordinates necessary.

Use distance/slope/midpoint formula.

Make a statement what the distance/slope/midpoint formula has shown.

Concept 4: Coordinate Proof

Given A(0, 0), B(p, q), and C(2p, 0), prove that

triangle ABC is isosceles.

Example 4

A(0, 0) C(2p, 0)

B(p, q)

𝐴𝐵 = (𝑝 − 0)2+(𝑞 − 0)2

𝐴𝐵 = 𝑝2 + 𝑞2

𝐵𝐶 = (2𝑝 − 𝑝)2+(0 − 𝑞)2

𝐵𝐶 = 𝑝2 + (−𝑞)2

𝐵𝐶 = 𝑝2 + 𝑞2

AC is horizontal because the y-coordinates are both 0, and thus AC=2p-0 AC=2p

Because AB=BC, triangle ABC has to be isosceles (by definition of an isosceles triangle.)

“Bad is never good until worse happens.” –Danish Proverb

Concept 5: Perpendicular Bisector

A perpendicular bisector is a segment, line, ray, or plane that is both perpendicular to a segment and intersects at the midpoint.

A B

m

A B

m

A B

m

A B

m

A B

m

Draw line CD that is a perpendicular bisector to

segment AB at point E. If AE=15, what is EB?

Example 1

A B

C

D

E

Concept 6: Perpendicular Bisector Theorem

In a plane, if a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment.

A B

C

D

If 𝐶𝐷 is the perpendicular bisector of 𝐴𝐵, then AC=BC.

Equidistant-A point that is the same distance from two different figures.

Find AC.

Example 2

A B

C

D

A

B

C

D

x 25

15

20

4x-5

2x+15

Concept 6: (Converse of the) Perpendicular Bisector Theorem

In a plane, if a point is equidistant from the endpoints of the segment, then it is on the perpendicular bisector of a segment.

A B

C

D

If 𝐶𝐷 is the perpendicular bisector of 𝐴𝐵 and AC=BC,

then E is on 𝐶𝐷.

E

Will line AB pass through point E?

Example 3

A

B

C D

E

A

B C D

E

A

B C D

E

Is line CD a perpendicular bisector of segment AB?

Example 4

A B

C

D

E

A B

C

D

E

A B

C

D

E

Concept 7: Lines being Concurrent

When three or more lines, rays, or segments intersect in the same point, they are called concurrent lines, rays, or segments.

“point of concurrency”

Concurrent lines Not concurrent lines

Concept 8: Point of Concurrency (Circumcenter)

Concurrency of Perpendicular Bisectors of a Triangle.

The perpendicular bisectors of a triangle intersect at a point that is equidistant from the vertices of the triangle.

Example 5

A

B

C

D E

F

G

Find the value of EG, AG, and BG if all the blue segments are perpendicular bisectors of a side.

15

20

16

You own a business that has 3 stores. The stores

require that you have a warehouse and you want one warehouse that is the same distance from all three stores. How would you find the location that is the same distance from each store?

Example 6

“Things could be worse. Suppose your errors were counted and published every day, like

those of a baseball player.” –Anon.

Concept 9: Angle Bisector

An angle bisector is a ray that divides (cuts) an angle into two congruent angles.

A

B

C

D

If ray BD bisects angle ABC, then ∠𝐴𝐵𝐷 ≅ ∠𝐷𝐵𝐶.

What is the measure of ∠ABC, if the m∠ABD=25°

and ray BD is an angle bisector of ∠ABC.

Example 1

A

B

C

D

m∠ABD=m∠DBC m∠DBC=25° m∠ABC=25°+25° m∠ABC=50°

Concept 10: Angle Bisector Theorem

If a point is on the bisector of an angle, then it is equidistant from the two sides of the angle.

A

B

C

D If

, then AD=DC.

Find the value of x.

Example 2

A

B

C

D 2x-4

X+2

Is AD=DC? Explain.

Example 3

A

B

C

D

A

B

C

D

Concept 10: Converse of the Angle Bisector Theorem

If a point is in the interior of an angle and is equidistant from the two sides of the angle, then it lies on the bisector of an angle.

A

B

C

D If

, then

A

B

C

D

Find the value of x.

Example 4

A

B

C

D (3x-5)°

(X+3)°

A

B

C

D 5x-4

3X+2

Concurrency of Angle Bisectors of a Triangle

The angle bisectors of a triangle intersect at a point that is equidistance from the sides of a triangle.

Concept 11: Incenter

Find the values of GF, GD, and FE, if G is the

incenter of the triangle.

Example 5

A

B

C

D E

F

G

25

20

“The mind of man(kind) is capable of anything – because everything is in it, all the

past as well as all the future.” –Joseph Conrad

Concept 12: Medians

A median is a segment that connects a vertex to the midpoint of the opposite side.

Is 𝐴𝐶 a median?

Example 1

A

B

C

D

A

B C D

A

B C D

A

B C

D

Concept 13: Centroid

Concurrency of Medians of a Triangle

The medians of a triangle intersect at a point that is two thirds of the distance from each vertex to the midpoint of the opposite side.

BG =2

3𝐵𝐹

AG =2

3𝐴𝐸

CG =2

3𝐶𝐷

BG = 2𝐺𝐹 AG = 2𝐺𝐸

CG = 2𝐺𝐷

FG =1

3𝐵𝐹

EG =1

3𝐴𝐸

DG =1

3𝐶𝐷

G is the centroid of ∆𝐴𝐵𝐶, BG=12, AF=8, AE=15, and DG=3. Find the length of the given segments.

Example 2

A

B C

D

E

F

G

12

8

3

BF=? FC=? GF=? AG=? GE=? GC=? DC=?

Find the coordinates of the centroid given the

coordinates of the vertices of the triangle. A(-1, 2), B(5, 6), and C(5, -2).

Example 3

Concept 14: Altitudes

The altitude of a triangle is the perpendicular segment from a vertex to the opposite side (or the line that contains the opposite side).

Isosceles Triangle-The altitude, median,

perpendicular bisector, and angle bisector are all the same segment from the vertex angle to base.

Equilateral/Equiangular Triangle-The altitude, median, perpendicular bisector, and angle bisector are all the same segment from any vertex to the opposite side.

Altitudes, Medians, Perpendicular Bisectors, and Angle Bisectors

Is 𝐴𝐶 a median, altitude, perpendicular bisectors?

Example 4

A

B

C

D

A

B C D

A

B C D

A

B C

D

Concept 15: Orthocenter

Concurrency of Altitudes of a Triangle

The lines containing the altitudes of a triangle are concurrent.

“That which is bitter to endure may be sweet to remember.”

–Thomas Fuller

Concept 16: Side Opposite

The two sides of a triangle that form an angle are adjacent to the angle. The side that is not adjacent is opposite to the angle.

A

B

C

Concept 17: Angles and Sides Match

Theorem 5.10

If one side of a triangle is longer than another side, then the angle opposite the longer side is larger than the angle opposite the shorter side.

A

B

C

If AB>AC, then m∠C>m∠B

Concept 17: Angles and Sides Match

Theorem 5.11

If one angle of a triangle is larger than another angle, then the side opposite the larger angle is larger than the side opposite the smaller angle.

A

B

C

If m∠C>m∠B, Then AB>AC.

Draw a sketch of the two triangles.

1) Side lengths: 5m, 3m, 7m; Angle measures: 22°, 38°, 120°

2) Side lengths: 6m, 9m, 10m; Angle measures: 81°, 63°, 36°

Example 1

3m 5m

7m

22° 38°

120° 6m 9m

10m

36° 63°

81°

Identify the triangle that are clearly impossible.

Example 2

3m 5m

7m

22°

38°

120°

3m

5m

7m

22°

68°

120°

3m 11m

7m

22°

69°

89°

4m 12m

8m

22°

69°

89°

3m 4m

1.5m

42°

28°

110°

3m 5m

7m

22° 38°

120°

Concept 18: Triangle Inequality Theorem

The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

A

B

C

BC+AB>AC

BC+AC>AB

AC+AB>BC

Identify the triangle that are clearly impossible.

Example 3

3m 5m

7m

22°

38°

120°

3m

5m

7m

22°

68°

120°

3m 11m

7m

22°

69°

89°

4m 12m

8m

22°

69°

89°

3m 4m

1.5m

42°

28°

110°

3m 5m

7m

22° 38°

120°

Is it possible to build a triangle using the given side

lengths?

1) 3m, 6m, 9m

2) 7m, 8m, 2m

3) 5m, 2m, 2m

4) 3m, 6m, 6m

5) 1m, 5m, 9m

6) 2m, 7m, 5m

7) 5m, 4m, 3m

Example 4

Explain why this diagram is impossible.

Example 5

15m

12m

5m

61°

60°

59° 59°

“Memory is the thing you forget with.” –Alexander Chase

If two sides of one triangle are congruent to two

sides of another triangle, and the included angle of the first is larger than the included angle of the second, then the third side of the first is longer than the third side of the second.

Concept 19: Hinge Theorem

A

B C

D

E F

25° 50°

, then EF>BC. If

If two sides of one triangle are congruent to two

sides of another triangle, and the third side of the first is longer than the third side of the second, then the included angle of the first is larger than the included angle of the second.

Concept 19: (Converse to the) Hinge Theorem

A

B C

D

E F

9m 10m

, then m∠D>m∠A. If

Determine which is larger.

Example 1

A

B C

D

E F

45°

35° EF___BC

A

B C

D

E F

5m

6m

m∠D___m∠A

Indirect reasoning involves making a temporary

assumption that the desired result is false.

Ex: I am trying to prove that all even numbers are divisible by two. I would make a temporary assumption that at least one even number is not divisible by 2.

Concept 20: Indirect Reasoning

Write a temporary assumption you could make to

prove the conclusion indirectly.

1) If x and y are even, then xy is even.

2) If xy is not zero, then x and y are both not zero.

3) If m∠A>m∠D, then BC>EF.

Example 2

The four steps to write an indirect proof.

1) Identify the statement we are trying to prove.

2) Assume temporarily that this statement is false by assuming that its opposite is true.

3) Reason normally until you reach a contradiction.

4) Point out that the desired conclusion must be true because the contradiction proves the assumption false.

Concept 21: Indirect Proof

Write an indirect proof of Theorem 5.11.

Given: m∠B>m∠C

Prove: AC>AB

Example 3

A

B C

Case 1: • Assume AC<AB. • If AC<AB, then m∠B<m∠C

by theorem 5.10. • This contradicts the given

and therefore AC<AB has to be false.

Case 2: • Assume AC=AB. • If AC=AB, then m∠B=m∠C

by the BA theorem. • This contradicts the given

and therefore AC=AB has to be false.

Therefore, AC>AB.