Let's add the midsegment, EF, to the diagram.
Note that O and C are on the x-axis which means they both have the y-coordinate. 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=pTherefore, F's coordinates are (p,0). To show that FE∥ OB, we have to prove that these segment have the same slope.
m_(EF)=m_(OB)
We already know the coordinates of O and B which means we can calculate the slope of OB. To determine the slope of EF, we need to figure out the coordinates of E. Since E is the midpoint of BC, we can find its coordinates by using the Midpoint Formula.
M(x_1+x_2/2,y_1+y_2/2)
M(2q+ 2p/2,2r+ 0/2)
M(2(q+p)/2,2r+0/2)
M(2(q+p)/2,2r/2)
M(q+p,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
|
| OB
|
( 2q,2r), ( 0,0)
|
2r- 0/2q- 0
|
r/q
|
| EF
|
( q+p,r), ( p,0)
|
r- 0/( q+p)- p
|
r/q
|
Both segments have a slope of rq which means they are parallel. Finally, we can show that FE= 12OB by calculating the length of FE and OB using the Distance Formula.
| Segment
|
Points
|
sqrt((x_2-x_1)^2+(y_2-y_1)^2)
|
d
|
| OB
|
( 2q,2r), ( 0,0)
|
sqrt(( 2q- 0)^2+( 2r- 0)^2)
|
sqrt(4q^2+4r^2)
|
| FE
|
( q+p,r), ( p,0)
|
sqrt((( q+p)- p)^2+( r- 0)^2)
|
sqrt(q^2+r^2)
|
By substituting the distances in the equation FE= 12OB, we can prove that the equation is true.
FE=1/2OB
sqrt(q^2+r^2)? =1/2( sqrt(4q^2+4r^2))
sqrt(q^2+r^2)? =1/2sqrt(4(q^2+r^2))
sqrt(q^2+r^2)? =1/2sqrt(4)sqrt(q^2+r^2)
sqrt(q^2+r^2)? =1/2* 2(sqrt(q^2+r^2))
sqrt(q^2+r^2)? =2/2sqrt(q^2+r^2)
sqrt(q^2+r^2)=sqrt(q^2+r^2)
As we can see, FE is half that of OB.
