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Transcript
GEOMETRY
MODULE 1 LESSON 29
SPECIAL LINES IN TRIANLGES
OPENING EXERCISE
A. Use your compass and straightedge to find the midpoints of Μ…Μ…Μ…Μ…
𝐴𝐡 and Μ…Μ…Μ…Μ…
𝐴𝐢 . Label the midpoint
as X and Y, respectively.
Μ…Μ…Μ…Μ….
B. Draw the midsegment π‘‹π‘Œ
The midsegment is a line segment that joins the midpoints of two sides of a triangle or
trapezoid.
ο‚·
Compare βˆ π΄π‘‹π‘Œ and ∠𝐴𝐡𝐢. Compare βˆ π΄π‘Œπ‘‹ and ∠𝐴𝐢𝐡. Without using a protractor, what
would you guess is the relationship between these two pairs of angles?
π‘šβˆ π΄π‘‹π‘Œ = π‘šβˆ π΄π΅πΆ
π‘šβˆ π΄π‘Œπ‘‹ = π‘šβˆ π΄πΆπ΅
ο‚·
Μ…Μ…Μ…Μ… ?
From your conclusion above, what can now be said about Μ…Μ…Μ…Μ…
π‘‹π‘Œ and 𝐡𝐢
Μ…Μ…Μ…Μ…
Μ…Μ…Μ…Μ… βˆ₯ 𝐡𝐢
π‘‹π‘Œ
HOMEWORK QUESTIONS
Problem Set Module 1 Lesson 28 is due Thursday.
MOD1 L29
1
DISCUSSION
Μ…Μ…Μ…Μ… and 𝐴𝐢
Μ…Μ…Μ…Μ… , we could have chosen any
Even though we chose to determine the midsegment of 𝐴𝐡
two sides to work with. Let’s now focus on the properties associated with a midsegment.
The midsegment of a triangle is parallel to the third side of the triangle and half the length of the
third side of the triangle.
Given: Μ…Μ…Μ…Μ…
π‘‹π‘Œ is a midsegment of βˆ†π΄π΅πΆ.
Μ…Μ…Μ…Μ… and π‘‹π‘Œ = 1 𝐡𝐢
Μ…Μ…Μ…Μ… βˆ₯ 𝐡𝐢
Prove: π‘‹π‘Œ
2
Μ…Μ…Μ…Μ… to point G so that π‘‹π‘Œ = π‘ŒπΊ. Draw 𝐺𝐢.
Construction: Extend π‘‹π‘Œ
STEP
JUSTIFICATION
1
Μ…Μ…Μ…Μ… is a midsegment of βˆ†π΄π΅πΆ
π‘‹π‘Œ
Given
2
π‘‹π‘Œ = π‘ŒπΊ
Construction
3
π‘šβˆ π΄π‘Œπ‘‹ = π‘šβˆ πΊπ‘ŒπΆ
Vertical Angles
4
π΄π‘Œ = π‘ŒπΆ
Property of Midpoint
5
βˆ†π΄π‘‹π‘Œ β‰… βˆ†πΆπΊπ‘Œ
SAS
6
𝐺𝐢 = 𝐴𝑋
Corresponding sides of congruent triangles are equal in length.
7
𝐺𝐢 = 𝐡𝑋
Substitution (Line 1: 𝐴𝑋 = 𝐡𝑋)
8
π‘šβˆ π΄π‘‹π‘Œ = π‘šβˆ πΆπΊπ‘Œ
Corresponding angles of congruent triangles are equal in measure.
9
Μ…Μ…Μ…Μ…
𝐴𝐡 βˆ₯ Μ…Μ…Μ…Μ…
𝐺𝐢
Alternate Interior Angles Converse
10
BXGC is a parallelogram
11
12
𝑋𝐺 = 𝐡𝐢
Μ…Μ…Μ…Μ…
Μ…Μ…Μ…Μ…
π‘‹π‘Œ βˆ₯ 𝐡𝐢
𝑋𝐺 = 2π‘‹π‘Œ
Substitution (Line #2 and π‘‹π‘Œ + π‘ŒπΊ = 𝑋𝐺)
13
𝐡𝐢 = 2π‘‹π‘Œ
Substitution (Line #11)
14
MOD1 L29
π‘‹π‘Œ =
1
𝐡𝐢
2
One pair of opposite sides is equal and parallel. (Line 7 and 9)
Property of Parallelogram
Division
2
PRACTICE
Apply what you know about the properties of midsegments to solve the following exercises.
1.
a. Find the length of x.
15
b. Find the perimeter of βˆ†π΄π΅πΆ.
68
2.
a. Find the angle measure of x.
50°
b. Find the angle measure of y.
70°
3. In βˆ†π‘…π‘†π‘‡, the midpoints of each side have been marked by points X, Y, and Z.
ο‚·
Mark the halves of each side divided by the midpoint with a
congruency mark.
ο‚·
Draw midsegments Μ…Μ…Μ…Μ…
π‘‹π‘Œ, Μ…Μ…Μ…Μ…
π‘Œπ‘, and Μ…Μ…Μ…Μ…
𝑋𝑍. March each
midsegment with the appropriate congruency mark for the
sides of the triangle.
a. What conclusion can you draw about the four triangles within βˆ†π‘…π‘†π‘‡? Why?
All four triangles are congruent by SSS.
b. State the appropriate correspondences among the four triangles within βˆ†π‘…π‘†π‘‡.
βˆ†π‘…π‘‹π‘Œ, βˆ†π‘Œπ‘π‘‡, βˆ†π‘‹π‘†π‘, βˆ†π‘π‘Œπ‘‹
c. State a correspondence between βˆ†π‘…π‘†π‘‡ and nay one of the four small triangles.
βˆ†π‘…π‘‹π‘Œ, βˆ†π‘…π‘†π‘‡
MOD1 L29
3
ON YOUR OWN
1. Find x.
Μ…Μ…Μ…Μ…Μ… and Μ…Μ…Μ…Μ…Μ…
2. R and S are the midpoints of π‘‹π‘Š
π‘Šπ‘Œ, respectively. Find the perimeter of βˆ†π‘Šπ‘‹π‘Œ.
SUMMARY
The midsegment of a triangle is parallel to the third side of the triangle and half the length of the
third side of the triangle.
1. 79° 2. 82
MOD1 L29
4