Question

triangle ( abc ) is shown with medians ( overleftrightarrow{ae} ) and ( overleftrightarrow{cd} ). line ( overleftrightarrow{ak} ) is drawn such that ( overleftrightarrow{ak} parallel overleftrightarrow{dc} ).

move options to the blanks to complete the proof that ( overline{bf} ) is the third median of triangle ( abc ).

1. because ( overleftrightarrow{ae} ) and ( overleftrightarrow{cd} ) are medians, ( overline{bd} cong ) ____ and ( overline{be} cong ) ____.
2. since ( overleftrightarrow{ak} parallel overleftrightarrow{dc} ), ( \frac{bd}{da} = \frac{bj}{jk} ).
3. 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 )
4. because ____ connects the midpoints of ( overline{bc} ) and ( overline{bk} ), __ ( parallel ) ____.
5. since it is also known that ( overleftrightarrow{dc} parallel overleftrightarrow{ak} ), ( ajck ) is a parallelogram.
6. the diagonals of a parallelogram bisect each other, so ____ ( cong ) ____ and ( overline{bf} ) is the third median of

options: ( overleftrightarrow{ae} ), ( overline{af} ), ( overline{bc} ), ( overline{bk} ), ( overline{da} ), ( overline{ec} ), ( overline{fc} )

Explanation:

Step1: Fill median segment congruences

A median connects a vertex to the midpoint of the opposite side, so $\overline{BD} \cong \overline{DA}$ and $\overline{BE} \cong \overline{EC}$.

Step2: Identify midpoint connecting segment

$\overline{JE}$ connects midpoints of $\overline{BC}$ and $\overline{BK}$, so $\overline{JE} \parallel \overline{AK}$.

Step3: Fill parallelogram diagonal congruence

Diagonals of parallelogram $AICK$ bisect each other, so $\overline{AF} \cong \overline{FC}$.

Answer:

1. $\overline{DA}$, $\overline{EC}$
2. $\overline{JE}$, $\overline{AK}$
3. $\overline{AF}$, $\overline{FC}$