Question

a. name the midsegments and sides that are parallel in $delta fhk$.b. name the midsegments that are half as long as each side of $delta fhk$.$square$ e. $overline{fh}||overline{jl}LXB0square$ b. $gj=\frac{1}{2}fkLXB1square$ d. $fl=\frac{1}{2}fkLXB2square$ f. there are no midsegments that are half as long as any

Explanation:

Step1: Identify midsegments (part a)

Midsegments connect midpoints: $\overline{GJ}$, $\overline{JL}$, $\overline{GL}$. By Triangle Midsegment Theorem, midsegments are parallel to a triangle side.

• $\overline{GJ} \parallel \overline{FK}$
• $\overline{JL} \parallel \overline{FH}$
• $\overline{GL} \parallel \overline{HK}$

From given options: $\boldsymbol{\overline{FH}\parallel\overline{JL}}$ (option E) and $\boldsymbol{\overline{FK}\parallel\overline{GL}}$ (option F) are valid.

Step2: Match midsegment lengths (part b)

Midsegment = $\frac{1}{2}$ length of parallel side:

• $\overline{GJ} = \frac{1}{2}\overline{FK}$ (FK=14, GJ=7)
• $\overline{JL} = \frac{1}{2}\overline{FH}$ (FH=10, JL=5)
• $\overline{GL} = \frac{1}{2}\overline{HK}$ (HK=12, GL=6)

These correspond to options A, B, E.

Answer:

Part a:

E. $\overline{FH}\parallel\overline{JL}$
F. $\overline{FK}\parallel\overline{GL}$

Part b:

A. $\text{GL}=\frac{1}{2}\text{HK}$
B. $\text{GJ}=\frac{1}{2}\text{FK}$
E. $\text{JL}=\frac{1}{2}\text{FH}$