a. name the segments and sides that are parallel in $ \triangle f h k $.
b. name the midsegments that are half as long as each side of $ \triangle f h k $
a $ \overline{f k} \| \overline{g j} $
b. $ \overline{h k}\|\overline{j l} $
c. $ \overline{f k}\|\overline{g l} $
d. $ \overline{h k}\|\overline{g l} $
e $ \overline{f h}\|\overline{g j} $
f. $ \overline{f h}\|\overline{j l} $
g there are no parallel segments.
b. name the midsegments that are half as long as each side of $ \triangle f h k $. select all that apply.
a. $ j l=\frac{1}{2} f h $
b. $ f l=\frac{1}{2} f k $
c. $ g j=\frac{1}{2} f k $
d. $ h j=\frac{1}{2} h k $
e $ g l=\frac{1}{2} h k $
f. there are no midsegments that are half as long as any sides of the $ \triangle a b c $

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Accepted Answer

# Explanation:
## Step1: Recall the mid - segment theorem
The mid - segment of a triangle is parallel to the third side and half its length.
## Step2: Analyze part (a)
- For $ \overline{FK}\parallel\overline{GJ}$: Since $G$ and $J$ are mid - points (by the lengths $HG = GF = 5$ and $HJ=JK = 6$), by the mid - segment theorem, $GJ$ is parallel to $FK$.
- For $ \overline{HK}\parallel\overline{GL}$: Since $G$ and $L$ are mid - points (by the lengths $HG = GF = 5$ and $FL = LK=7$), by the mid - segment theorem, $GL$ is parallel to $HK$.
- For $ \overline{FH}\parallel\overline{JL}$: Since $J$ and $L$ are mid - points (by the lengths $HJ = JK = 6$ and $FL = LK = 7$), by the mid - segment theorem, $JL$ is parallel to $FH$.
## Step3: Analyze part (b)
- For $GJ=\frac{1}{2}FK$: Since $GJ$ is a mid - segment with respect to $FK$ ( $G$ is the mid - point of $FH$ and $J$ is the mid - point of $HK$), by the mid - segment theorem $GJ=\frac{1}{2}FK$.
- For $JL=\frac{1}{2}FH$: Since $JL$ is a mid - segment with respect to $FH$ ( $J$ is the mid - point of $HK$ and $L$ is the mid - point of $FK$), by the mid - segment theorem $JL=\frac{1}{2}FH$.
- For $GL=\frac{1}{2}HK$: Since $GL$ is a mid - segment with respect to $HK$ ( $G$ is the mid - point of $FH$ and $L$ is the mid - point of $FK$), by the mid - segment theorem $GL=\frac{1}{2}HK$.

# Answer:
a. A. $ \overline{FK}\parallel\overline{GJ}$, D. $ \overline{HK}\parallel\overline{GL}$, F. $ \overline{FH}\parallel\overline{JL}$
b. A. $JL=\frac{1}{2}FH$, C. $GJ=\frac{1}{2}FK$, E. $GL=\frac{1}{2}HK$

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QUESTION IMAGE

Question

Explanation:

Step1: Recall the mid - segment theorem

Step2: Analyze part (a)

Step3: Analyze part (b)

Answer: