Download Lesson 5-2 Perpendicular and Angle Bisectors

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Steinitz's theorem wikipedia , lookup

Cartesian coordinate system wikipedia , lookup

Rational trigonometry wikipedia , lookup

Atiyah–Singer index theorem wikipedia , lookup

Perceived visual angle wikipedia , lookup

Integer triangle wikipedia , lookup

Triangle wikipedia , lookup

Trigonometric functions wikipedia , lookup

3-manifold wikipedia , lookup

Riemann–Roch theorem wikipedia , lookup

History of trigonometry wikipedia , lookup

Noether's theorem wikipedia , lookup

Euclidean geometry wikipedia , lookup

Four color theorem wikipedia , lookup

Brouwer fixed-point theorem wikipedia , lookup

Pythagorean theorem wikipedia , lookup

Transcript
Lesson 5-2: Perpendicular and Angle
Bisectors
Goal: Use angle bisectors and perpendicular
bisectors to prove triangles congruent or find
missing measures.
Key Words:
Theorem 5.3:
Example 1:
Use the Angle Bisector Theorem
 In the diagram, 𝐸𝐺 bisects ∠𝐶𝐸𝐹.
Prove that ∆𝐶𝐷𝐸 ≅ ∆𝐹𝐷𝐸.
Statements
Reasons
1. 𝐸𝐺 bisects ∠𝐶𝐸𝐹
Given
2. ∠𝐶𝐸𝐷 ≅ ∠𝐹𝐸𝐷
Def. of angle bisector
3. ∠𝐷𝐶𝐸 𝑎𝑛𝑑∠𝐷𝐹𝐸 are right ∠′ 𝑠
Given
4. ∠𝐷𝐶𝐸 ≅ ∠𝐷𝐹𝐸
All right angles are ≅.
5. 𝐷𝐶 ≅ 𝐷𝐹
Angle Bisector Theorem
6. ∆𝐶𝐷𝐸 ≅ ∆𝐹𝐷𝐸
AAS
Theorem 5.4:
Example 2: Use Perpendicular Bisectors
 Use the diagram to find GH.
3𝑥 + 2 = 5𝑥 − 12
2 = 2𝑥 − 12
14 = 2𝑥
14
2
=
2𝑥
2
7=𝑥
by the Perpendicular Bisector Theorem GH = KH.
Subtract 3x from both sides.
Subtract 12 from both sides.
Divide both sides by 2.
Simplify.
𝐺𝐻 = 3 7 + 2 = 21 + 2 = 23
Your Turn:
𝑥 + 3 = 2𝑥 + 1
4𝑥 = 𝑥 + 15
2𝑥 + 5 = 𝑥 + 10
2=𝑥
3𝑥 = 15
𝑥=5
𝐹𝐻 = 2 + 3 = 5
𝑥=5
𝐸𝐹 = 2 5 + 5 = 15
𝑀𝐾 = 5 + 15 = 20
Example 3: Using the Perpendicular
Bisector Theorem
 Can you prove that ∆𝑀𝑆𝑇 is an isosceles triangle if 𝑀𝑁 is
only known to be a bisector of 𝑆𝑇.
No, if 𝑀𝑁 is not known to be
perpendicular we can not conclude
that 𝑀𝑇 and 𝑀𝑆 are congruent.
Checkpoint:
equidistant
Perpendicular bisector
𝒙 + 𝟏 = 𝟐𝒙 − 𝟏
16
𝟐=𝒙
𝑬𝑭 = 𝟐 + 𝟏 = 𝟑
𝟓𝒙 = 𝟑𝒙 + 𝟖
12
𝒙=𝟒
𝑸𝑹 = 𝟓(𝟒) = 𝟐𝟎
Assignment:
 p 296 – 298 #’s 6 – 34 even