Question

given: abcd is a rhombus;
\\( \overline { be } \cong \overline { cf } ; \overline { ce } \cong \overline { fd } \\)
prove: abcd is a square.

1. abcd is a rhombus.
2. \\( \overline { a b } \cong \overline { b c } \cong \overline { c d } \cong \overline { a d } \\)
3. \\( \overline { b e } \cong \overline { c f } , \overline { c e } \cong \overline { f d } \\)
4. \\( \triangle c b e \cong \triangle d c f \\)

Explanation:

Step1: Properties of a rhombus

In a rhombus, all sides are equal. So, if \(ABCD\) is a rhombus, then \(\overline{AB}\cong\overline{BC}\cong\overline{CD}\cong\overline{AD}\). The reason for this statement is the definition of a rhombus.

Step2: Given information

We are given that \(\overline{BE}\cong\overline{CF}\) and \(\overline{CE}\cong\overline{FD}\). The reason for this is "Given" as it is provided in the problem statement.

Step3: Proving triangles congruent

For \(\triangle CBE\) and \(\triangle DCF\):

• We know \(\overline{BC}\cong\overline{CD}\) (from the property of the rhombus, all sides of a rhombus are equal).
• \(\overline{BE}\cong\overline{CF}\) (given).
• \(\overline{CE}\cong\overline{FD}\) (given).

By the Side - Side - Side (SSS) congruence criterion, \(\triangle CBE\cong\triangle DCF\). The reason for \(\triangle CBE\cong\triangle DCF\) is the SSS (Side - Side - Side) congruence postulate.

Answer:

1. Definition of a rhombus; 3. Given; 4. SSS (Side - Side - Side) congruence postulate.