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Big Ideas Math Geometry, 2014

BI

Big Ideas Math Geometry, 2014 View details

4. The Triangle Midsegment Theorem

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Exercise 22 Page 334

What characteristics do parallel lines have in a coordinate plane? How can you calculate the length of a segment?

See solution.

Practice makes perfect

Let's add the midsegment, DF, to the diagram.

To prove the Triangle Midsegment Theorem, we have to show that DF∥ BC and that DF= 12BC.

Proving DF∥ BC

If DF∥ BC, these segments will have the same slope. To find the slope of a segment, we need the coordinates of the endpoints. We already know the coordinates of BC, but we also need to figure out the coordinates of F and D.
Note that O and C are on the x-axis which means they both have the y-coordinate 0. Therefore, the midpoint F must also have the y-coordinate 0. To calculate F's x-coordinate, we add the x-coordinates of O and C and divide by 2. x-coordinate F: 0+ 2p/2=p Therefore, F's coordinates are (p,0). To determine the coordinates of D we can use the Midpoint Formula.

M(x_1+x_2/2,y_1+y_2/2)

SubstitutePoints

Substitute ( 2q,2r) & ( 0,0)

M(2q+ 0/2,2r+ 0/2)

▼

Simplify

Add terms

M(2q/2,2r/2)

Simplify quotient

M(q,r)

Now we can calculate the slope of the segments using the Slope Formula.

Segment	Points	y_2-y_1/x_2-x_1	m
BC	( 2q,2r), ( 2p,0)	2r- 0/2q- 2p	r/q-p
DF	( q,r), ( p,0)	r- 0/q- p	r/q-p

Both segments have a slope of rq-p which means they are parallel.

Proving BC= 12DF

Finally, we can show that BC= 12DF by calculating the lengths of BC and DF using the Distance Formula.

Segment	Points	sqrt((x_2-x_1)^2+(y_2-y_1)^2)	d
BC	( 2q,2r), ( 2p,0)	sqrt(( 2q- 2p)^2+( 2r- 0)^2)	sqrt((2q-2p)^2+4r^2)
DF	( q,r), ( p,0)	sqrt(( q- p)^2+( r- 0)^2)	sqrt((q-r)^2+r^2)

By substituting the distances in the equation DF= 12BC, we can prove that the equation is true.

DF=1/2BC

DF= sqrt((q-r)^2+r^2), BC= sqrt((2q-2p)^2+4r^2)

sqrt((q-r)^2+r^2)? =1/2( sqrt((2q-2p)^2+4r^2))

▼

Simplify right-hand side

Factor out 2

sqrt((q-r)^2+r^2)? =1/2sqrt((2(q-p))^2+4r^2)

(a * b)^m=a^m* b^m

sqrt((q-r)^2+r^2)? =1/2sqrt(4(q-p)^2+4r^2)

Factor out 4

sqrt((q-r)^2+r^2)? =1/2sqrt(4((q-p)^2+r^2))

sqrt(a* b)=sqrt(a)*sqrt(b)

sqrt((q-r)^2+r^2)? =1/2sqrt(4)sqrt((q-p)^2+r^2)

Calculate root

sqrt((q-r)^2+r^2)? =1/2* 2sqrt((q-p)^2+r^2)

MoveRightFacToNumOne

1/b* a = a/b

sqrt((q-r)^2+r^2)? =2/2sqrt((q-p)^2+r^2)

a/a=1

sqrt((q-r)^2+r^2)=sqrt((q-p)^2+r^2)

As we can see, DF is half that of BC.

Having proven that DF∥ BC and DF= 12BC, we know the Triangle Midsegment Theorem is true.

Subchapter links

Communicate Your Answer p.329

Monitoring Progress p.330-332