Let's add the midsegment, $\overline{E F},$ 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 - c o o r d i n a t e F : \frac{0 + 2 p}{2} = p

Therefore,

F

's coordinates are

(p, 0) .

To show that

\overline{F E} ∥ \overline{O B},

we have to prove that these segment have the same slope.

m_{E F} = m_{O B}

We already know the coordinates of

O

and

B

which means we can calculate the slope of

\overline{O B} .

To determine the slope of

\overline{E F},

we need to figure out the coordinates of

E .

Since

E

is the midpoint of

\overline{B C},

we can find its coordinates by using the Midpoint Formula.

M (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})

SubstitutePoints

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

M (\frac{2 q + 2 p}{2}, \frac{2 r + 0}{2})

▼

Simplify

FactorOut

Factor out $2$

M (\frac{2 ( q + p )}{2}, \frac{2 r + 0}{2})

AddTerms

Add terms

M (\frac{2 ( q + p )}{2}, \frac{2 r}{2})

SimpQuot

Simplify quotient

M (q + p, r)

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

Segment	Points	$\frac{y _{2} - y _{1}}{x _{2} - x _{1}}$	$m$
$\overline{O B}$	$(2 q, 2 r), (0, 0)$	$\frac{2 r - 0}{2 q - 0}$	$\frac{r}{q}$
$\overline{E F}$	$(q + p, r), (p, 0)$	$\frac{r - 0}{( q + p ) - p}$	$\frac{r}{q}$

Both segments have a slope of $\frac{r}{q}$ which means they are parallel. Finally, we can show that $F E = \frac{1}{2} O B$ by calculating the length of $\overline{F E}$ and $\overline{O B}$ using the Distance Formula.

Segment	Points	$(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$	$d$
$\overline{O B}$	$(2 q, 2 r), (0, 0)$	$(2 q - 0)^{2} + (2 r - 0)^{2}$	$4 q^{2} + 4 r^{2}$
$\overline{F E}$	$(q + p, r), (p, 0)$	$((q + p) - p)^{2} + (r - 0)^{2}$	$q^{2} + r^{2}$

Exercise

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