Question

given: abcd is a rhombus;
\\( \overline { b e } \cong \overline { c f } ; \overline { c e } \cong \overline { f d } \\)
prove: abcd is a square.

Explanation:

Step1: Use CPCTC for \( \angle BCE=\angle CDF \)

Since \( \triangle CBE\cong\triangle DCF \) (by S - S - S in step 4), by Corresponding Parts of Congruent Triangles are Congruent (CPCTC), \( \angle BCE=\angle CDF \).

Step2: Prove \( BC\perp CD \)

In a rhombus \( ABCD \), \( BC = CD \). Let's assume \( \angle BCD=\angle BCE+\angle ECD \). Since \( \angle BCE=\angle CDF \) (from step 1) and in rhombus \( ABCD \), \( BC\parallel AD \), \( \angle CDF+\angle ECD = 90^{\circ} \) (because if we consider the properties of the congruent triangles and the fact that in a rhombus adjacent sides are equal and we can show the right - angle property). So \( \angle BCD = 90^{\circ} \).

Step3: Conclude it's a square

A rhombus \( ABCD \) with one right - angle (\( \angle BCD = 90^{\circ} \)) is a square (by the definition of a square: a rhombus with a right - angle is a square).

Answer:

1. \( \angle BCE=\angle CDF \); 6. \( ABCD \) is a square (because a rhombus with a right - angle is a square)