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.

1. \\( m \angle b + m \angle d c f = 180 \\)
2. \\( m \angle b + m \angle b = 180 \\) or \\( 2 m \angle b = 180 \\)
3. \\( m \angle b = 90 \\)
4. if two parallel lines

are cut by a transver-
sal, then their interior
angles on the same
side of the transversal
are supplementary.

1. click here to insert
2. click here to insert

Explanation:

Step1: Substitute $\angle DCF$ with $\angle B$

Since $ABCD$ is a rhombus, $AB\parallel CD$. And from $\overline{BE}\cong\overline{CF}$ and $\overline{CE}\cong\overline{FD}$, we can prove $\triangle BCE\cong\triangle CDF$ (by SSS congruence criterion). So, $\angle B=\angle DCF$. Then in the equation $m\angle B + m\angle DCF=180$, we substitute $\angle DCF$ with $\angle B$.

Step2: Solve for $m\angle B$

We have $m\angle B + m\angle B = 180$ (from substitution in Step 1), which simplifies to $2m\angle B=180$. Dividing both sides by 2, we get $m\angle B=\frac{180}{2}=90$.

Answer:

for 8:
Substitution (since $\angle B=\angle DCF$ from congruent triangles and parallel - side properties of the rhombus)