QUESTION IMAGE
Question
move options to the blanks to complete the proof that \\(\overline{bf}\\) is the third median of triangle \\(abc\\).
- because \\(\overline{ae}\\) and \\(\overline{cd}\\) are medians, \\(\overline{bd} \cong \underline{\overline{da}}\\) and \\(\overline{be} \cong \underline{\overline{bc}}\\).
- since \\(\overline{ak} \parallel \overline{dc}\\), \\(\frac{bd}{da} = \frac{bj}{jk}\\).
- using substitution and the fact that \\(\overline{bd} \cong \overline{da}\\), \\(1 = \frac{bj}{jk}\\). this shows that \\(\overline{bj} \cong \overline{jk}\\), which makes \\(j\\) the m
- because __ connects the midpoints of \\(\overline{bc}\\) and \\(\overline{bk}\\), \\(\parallel\\) __.
- since it is also known that \\(\overline{dc} \parallel \overline{ak}\\), \\(ajck\\) is a parallelogram.
- the diagonals of a parallelogram bisect each other, so __ \\(\cong\\) __ and \\(\overline{bf}\\) is the third median of the
options: \\(\overline{ae}\\), \\(\overline{af}\\), \\(\overline{bc}\\), \\(\overline{bk}\\), \\(\overline{da}\\), \\(\overline{ec}\\), \\(\overline{fc}\\), \\(\overline{fk}\\)
Step1: Fix median segment error
Median $\overline{AE}$ means $\overline{BE} \cong \overline{EC}$ (not $\overline{BC}$).
Step2: Identify midpoint connector
$\overline{JE}$ connects midpoints of $\overline{BC}$ and $\overline{BK}$, so $\overline{JE} \parallel \overline{CK}$.
Step3: Use substitution for ratio
Since $\overline{BD} \cong \overline{DA}$, $\frac{BD}{DA}=1$, so $1=\frac{BJ}{JK}$, meaning $\overline{BJ} \cong \overline{JK}$.
Step4: Apply parallelogram diagonal rule
Diagonals of parallelogram $AJCK$ bisect each other, so $\overline{FC} \cong \overline{AF}$.
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- Corrected blank: $\boldsymbol{\overline{EC}}$ (for $\overline{BE} \cong \boldsymbol{\overline{EC}}$)
- Blank 4: $\boldsymbol{\overline{JE}}$
- Blank 5: $\boldsymbol{\overline{JE}} \parallel \boldsymbol{\overline{CK}}$
- Blank 6: $\boldsymbol{\overline{AF}} \cong \boldsymbol{\overline{FC}}$