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Transcript
Geometry- Lesson 9
Unknown Angle Proofs- Writing
Proofs
1
Essential Question
β€’ Write unknown angle proofs, which does not
require any new geometric facts.
β€’ String together facts you already know to
reveal more information.
2
Opening Exercise (5 min)
One of the main goals in studying geometry is to
develop your ability to reason critically, to draw
valid conclusions based upon observations and
proven facts. Master detectives do this sort of thing
all the time. Take a look as Sherlock Holmes uses
seemingly insignificant observations to draw
amazing conclusions.
Sherlock Holmes: Master of Deduction!
Could you follow Sherlock Holmes’s reasoning as he described
his thought process?
3
Discussion (10 min)
You needed to figure out the
measure of 𝒂, and used the
β€œfact” that an exterior angle of a
triangle equals the sum of the
measures of the opposite
interior angles. The measure of
𝒂 must therefore be πŸ‘πŸ”°.
Suppose that we rearrange the diagram just a little bit.
Instead of using numbers we’ll use variables to represent angle
measures…
4
Suppose further that we already have in our arsenal of
facts the knowledge that the angles of a triangle sum to
180°. Given the labeled diagram at the right, can we prove
that π‘₯ + 𝑦 = 𝑧 (or, in other words, that the exterior angle of
a triangle equals the sum of the remote interior angles)?
Proof:
Label βˆ π‘€, as shown in the diagram.
∠π‘₯ + βˆ π‘¦ + βˆ π‘€ = 180˚
βˆ π‘€ + βˆ π‘§ = 180˚
∠π‘₯ + βˆ π‘¦ + βˆ π‘€ = βˆ π‘€ + βˆ π‘§
∠ sum of β–³
∠s on a line
∴ ∠π‘₯ + βˆ π‘¦ = βˆ π‘§
5
Notice that each step in the proof was justified by a previously
known or demonstrated fact.
We ended up with a newly proven fact:
An exterior angle of any triangle is the sum of the remote
interior angles of the triangle.
∠π‘₯ + βˆ π‘¦ = βˆ π‘§
This ability to identify the steps used to reach a conclusion based
on known facts is deductive reasoning. The same type of
reasoning that Sherlock Holmes used to accurately describe the
doctor’s attacker in the video clip!
6
Exercises (25 min)
1. You know that angles on a line sum to 180°.
Prove that vertical angles are congruent.
Make a plan:
β€’ What do you know about βˆ π‘€ and ∠π‘₯
βˆ π‘¦ and ∠π‘₯
They sum to πŸπŸ–πŸŽ°.
β€’ What conclusion can you draw
based on both bits of knowledge?
βˆ π’˜=βˆ π’š
7
Write out your proof:
1.)
∠ π’˜ + ∠ 𝒙 = πŸπŸ–πŸŽ°
2.)
∠ π’š + ∠ 𝒙=πŸπŸ–πŸŽ°
3.)
∠ π’˜ + ∠ 𝒙= ∠ π’š + ∠ 𝒙
∴
∠ π’˜= ∠ π’š
8
2. Given the diagram at the right, prove that
∠ π’˜ + ∠ 𝒙 + ∠ 𝒛 = πŸπŸ–πŸŽ°.
(Make a plan first. What do you know about ∠ 𝒙, ∠ π’š, and
∠ 𝒛?)
Statement
∠ π’š + ∠ 𝒙 + ∠ 𝒛 = πŸπŸ–πŸŽ°
βˆ π’š=βˆ π’˜
Reason
∠ sum of a β–³
vert. ∠ s
∴ ∠ π’˜ + ∠ 𝒙 + ∠ 𝒛 = πŸπŸ–πŸŽ°
9
3. Given the diagram at the right, prove that
∠ π’˜ = ∠ π’š + ∠ 𝒛.
Statement
βˆ π’˜=βˆ π’™+βˆ π’›
βˆ π’™+βˆ π’š
Reason
ext. ∠ of a β–³
vert. ∠ s
βˆ΄βˆ π’˜=βˆ π’š+βˆ π’›
10
4. In the diagram at the right, prove that
∠ π’š + ∠ 𝒛 = ∠ π’˜ + ∠ 𝒙.
(You will need to write in a label in the diagram that is not
labeled yet for this proof.)
Statement
∠ 𝒂 + ∠ 𝒙 + ∠ π’˜= πŸπŸ–πŸŽ°
∠ 𝒂 + ∠ 𝒛 + ∠ π’š= πŸπŸ–πŸŽ°
Reason
ext. ∠ of a β–³
ext. ∠ of a β–³
βˆ π’‚+βˆ π’™+βˆ π’˜=βˆ π’‚=βˆ π’›+βˆ π’š
βˆ΄βˆ π’™+βˆ π’˜=βˆ π’›+βˆ π’š
a
11
5. In the figure at the right, 𝑨𝑩 || π‘ͺ𝑫 and 𝑩π‘ͺ || 𝑫𝑬.
Prove that ∠ 𝑨𝑩π‘ͺ = ∠ π‘ͺ𝑫𝑬.
Statement
Reason
∠ 𝑨𝑩π‘ͺ = ∠ 𝑫π‘ͺ𝑩
alt. ∠ s
∠ 𝑫π‘ͺ𝑩 = ∠ π‘ͺ𝑫𝑬
alt. ∠ s
∴ ∠ 𝑨𝑩π‘ͺ = ∠ π‘ͺ𝑫𝑬
12
6. In the figure at the right, prove that the sum of the
angles marked by arrows is 900°.
(You will need to write in several labels into the diagram for
this proof.)
Statement
𝒂 + 𝒃 + 𝒄 + 𝒅 = πŸ‘πŸ”πŸŽ°
𝒆 + 𝒇 + π’ˆ + 𝒉 = πŸ‘πŸ”πŸŽ°
Reason
f
∠ s at a pt.
e
∠ s at a pt.
h
a
π’Š + 𝒋 + π’Œ + π’Ž = πŸ‘πŸ”πŸŽ°
∠ s at a pt.
𝒂+𝒃+𝒄+𝒅+𝒆+𝒇+π’ˆ+𝒉+π’Š+π’Œ+π’Ž=πŸπŸŽπŸ–πŸŽ°
𝒅 + 𝒉 + π’Š = πŸπŸ–πŸŽ°
b
g
d
c
j
i
m
k
∠ sum of a β–³
∴ 𝒂+𝒃+𝒄+𝒆+𝒇+π’ˆ+𝒋+π’Œ+π’Ž = πŸ—πŸŽπŸŽ°
13
7. In the figure at the right, prove that 𝑫π‘ͺ βŠ₯ 𝑬𝑭.
Draw in label 𝒁.
𝒁
Statement
∠ 𝑬 + ∠ 𝑨 + ∠ 𝑬𝑭𝑨 = πŸπŸ–πŸŽ°
Reason
∠ sum of a β–³
∠ 𝑬𝑭𝑨 = πŸ”πŸŽΛš
∠ 𝑩 + ∠ π‘ͺ + ∠ C𝑫𝑩 = πŸπŸ–πŸŽ°
∠ π‘ͺ𝑫𝑩 = πŸ‘πŸŽ°
∠ sum of a β–³
∠ π‘ͺ𝑫𝑩 + ∠ 𝑬𝑭𝑨 + ∠ DZF = πŸπŸ–πŸŽ° ∠ sum of a β–³
∠ DZF = πŸ—πŸŽΛš
defn. of perpendicular
𝑫π‘ͺβŠ₯𝑬𝑭
14
Exit Ticket
In the diagram at the right, prove that the sum of the labeled
angles is 180°.
Statement
Reason
Label ∠ π’˜
vertical to ∠ x
βˆ π’˜=βˆ π’™
vert. ∠ s
∠ 𝒛 + ∠ π’˜ + ∠ π’š = πŸπŸ–πŸŽ°
∠ s on a line
∠ 𝒛 + ∠ 𝒙 + ∠ π’š = πŸπŸ–πŸŽ°
15