Question

given: abcd is a rhombus and \\(\overline{df}\\) bisects \\(\overline{bc}\\).
prove: \\(\overline{bf} \cong \overline{ad}\\).

step | statement | reason
1 | abcd is a rhombus; \\(\overline{df}\\) bisects \\(\overline{bc}\\) | given
2 | \\(\overline{be} \cong \overline{ce}\\) | a segment bisector divides a segment into two congruent segments
3 | \\(\angle dec \cong \angle feb\\) | vertical angles are congruent
4 | \\(\overline{dc} \parallel \overline{ab}\\) | opposite sides of a parallelogram / rhombus / rectangle / square are parallel
5 | \\(\angle c \cong \angle ebf\\) | parallel lines cut by a transversal form congruent alternate interior angles
6 | \\(\triangle dec \cong \triangle feb\\) | asa
7 | \\(\overline{bf} \cong \overline{dc}\\) | corresponding parts of congruent triangles are congruent (cpctc)
8 | \\(\overline{dc} \cong \overline{ad}\\) | all sides of a rhombus / square are congruent
9 | \\(\overline{bf} \cong \overline{ad}\\) | to be filled

Explanation:

Step 1: Recall the property of congruence (transitive property)

If \( \overline{BF} \cong \overline{DC} \) (from step 7) and \( \overline{DC} \cong \overline{AD} \) (from step 8), then by the transitive property of congruence (if \( a \cong b \) and \( b \cong c \), then \( a \cong c \)), we can conclude \( \overline{BF} \cong \overline{AD} \).

Step 2: Apply the transitive property

Let \( a=\overline{BF} \), \( b = \overline{DC} \), \( c=\overline{AD} \). Since \( a\cong b \) and \( b\cong c \), by transitive property of congruent segments, \( a\cong c \), i.e., \( \overline{BF} \cong \overline{AD} \). The reason is "Transitive Property of Congruence (If two segments are congruent to the same segment, then they are congruent to each other)".

Answer:

The reason for step 9 is "Transitive Property of Congruence (If two segments are congruent to the same segment, then they are congruent to each other)" and we have proven \( \overline{BF} \cong \overline{AD} \).