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5.4 Midsegment Theorem. Geometry Mrs. Spitz Fall 2004. Objectives:. Identify the midsegments of a triangle. Use properties of midsegments of a triangle. Assignment. Pgs. 290-291 #1-18, 21-22, 26-29. Using Midsegments of a Triangle.

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5 4 midsegment theorem

5.4 Midsegment Theorem

Geometry

Mrs. Spitz

Fall 2004

objectives
Objectives:
  • Identify the midsegments of a triangle.
  • Use properties of midsegments of a triangle.
assignment
Assignment
  • Pgs. 290-291 #1-18, 21-22, 26-29
using midsegments of a triangle
Using Midsegments of a Triangle
  • In lessons 5.2 and 5.3, you studied four special types of segments of a triangle:
    • Perpendicular bisectors
    • Angle bisectors
    • Medians and
    • Altitudes
  • A midsegment of a triangle is a segment that connects the midpoints of two sides of a triangle.
slide5
How?
  • You can form the three midsegments of a triangle by tracing the triangle on paper, cutting it out, and folding it as shown.
  • Fold one vertex onto another to find one midpoint.
  • Repeat the process to find the other two midpoints.
  • Fold a segment that contains two of the midpoints.
  • Fold the remaining two midsegments of the triangle.
solution
M= -2+6 , 3+(-1)

2 2

M = (2, 1)

And

N = 4+6 , 5+(-1)

2 2

N = (5, 2)

Next find the slopes of JK and MN.

Slope of JK = 5 – 3 = 2 = 1

4-(-2) 6 3

Slope of MN= 2 – 1 = 1

5 – 2 3

►Because their slopes are equal, JK and MN are parallel. You can use the Distance Formula to show that MN = √10 and JK = √40 = 2√10. So MN is half as long as JK.

Solution:
theorem 5 9 midsegment theorem
The segment connecting the midpoints of two sides of a triangle is parallel to the third side and is half as long.

DE ║ AB, and

DE = ½ AB

Theorem 5.9: Midsegment Theorem
ex 2 using the midsegment theorem
UW and VW are midsegments of ∆RST. Find UW and RT.

SOLUTION:

UW = ½(RS) = ½ (12) = 6

RT = 2(VW) = 2(8) = 16

A coordinate proof of Theorem 5.9 for one midsegment of a triangle is given on the next slide. Exercises 23-25 ask for proofs about the other two midsegments. To set up a coordinate proof, remember to place the figure in a convenient location.

Ex. 2: Using the Midsegment Theorem
ex 3 proving theorem 5 9
Write a coordinate proof of the Midsegment Theorem.

Place points A, B, and C in convenient locations in a coordinate plane, as shown. Use the Midpoint formula to find the coordinate of midpoints D and E.

Ex. 3: Proving Theorem 5.9
ex 3 proving theorem 5 911
Ex. 3: Proving Theorem 5.9

D = 2a + 0 , 2b + 0 = a, b

2 2

E = 2a + 2c , 2b + 0 = a+c, b

2 2

Find the slope of midsegment DE. Points D and E have the same y-coordinates, so the slope of DE is 0.

►AB also has a slope of 0, so the slopes are equal and DE and AB are parallel.

now what
Now what?

Calculate the lengths of DE and AB. The segments are both horizontal, so their lengths are given by the absolute values of the differences of their x-coordinates.

AB = |2c – 0| = 2c DE = |a + c – a | = c

►The length of DE is half the length of AB.

objective 2 using properties of midsegments
Objective 2: Using properties of Midsegments
  • Suppose you are given only the three midpoints of the sides of a triangle. Is it possible to draw the original triangle? Example 4 shows one method.
slide14
What?

PLOT the midpoints on the coordinate plane.

CONNECT these midpoints to form the midsegments LN, MN, and ML.

FIND the slopes of the midsegments. Use the slope formula as shown.

slide15
What?

ML = 3-2 = -1

2-4 2

MN = 4-3 = 1

5-2 3

LN = 4-2 = 2 = 2

5-4 1

Each midsegment contains two of the unknown triangle’s midpoints and is parallel to the side that contains the third midpoint. So you know a point on each side of the triangle and the slope of each side.

slide16
What?

►DRAW the lines that contain the three sides.

The lines intersect at 3 different points.

A (3, 5)

B (7, 3)

C (1, 1)

The perimeter formed by the three midsegments of a triangle is half the perimeter of the original triangle shown in example #5.

ex 5 perimeter of midsegment triangle
DF = ½ AB = ½ (10) = 5

EF = ½ AC = ½ (10) = 5

ED = ½ BC=½ (14.2)= 7.1

►The perimeter of ∆DEF is 5 + 5 + 7.1, or 17.1. The perimeter of ∆ABC is 10 + 10 + 14.2, or 34.2, so the perimeter of the triangle formed by the midsegments is half the perimeter of the original triangle.

Ex. 5: Perimeter of Midsegment Triangle