Question

vera wants to prove that any rectangle is also a parallelogram.
rectangle abcd with vertices labeled a, b, c, d
select the appropriate rephrased statement for veras proof.
choose 1 answer:
a) quadrilateral abcd, if \\(\overline{ad} \parallel \overline{bc}\\), then \\(m\angle a = m\angle b = m\angle c = \dots\\)
b) quadrilateral abcd, if \\(m\angle a = m\angle b = m\angle c = \dots\\), then \\(\overline{ab} \parallel \overline{dc}\\) and \\(\overline{ad} \parallel \dots\\)
c) quadrilateral abcd, if \\(\overline{ad} \cong \overline{bc}\\), then \\(m\angle a = m\angle b = m\angle c = \dots\\)
d) quadrilateral abcd, if \\(m\angle a = m\angle b = m\angle c = \dots\\), then \\(\overline{ab} \cong \overline{dc}\\) and \\(\overline{ad} \cong \dots\\)

Explanation:

To prove a rectangle is a parallelogram, we use the properties of rectangles (all angles \(90^\circ\)) and parallelograms (opposite sides parallel). A rectangle has all angles equal (\(m\angle A = m\angle B = m\angle C = m\angle D = 90^\circ\)). For a quadrilateral to be a parallelogram, opposite sides must be parallel. Option B states: "For quadrilateral \(ABCD\), if \(m\angle A = m\angle B = m\angle C = m\angle D = 90^\circ\), then \(\overline{AB} \parallel \overline{DC}\) and \(\overline{AD} \parallel \overline{BC}\)", which aligns with proving a rectangle (with all right angles) has opposite sides parallel (defining a parallelogram). Other options either use incorrect conditions (like side congruence instead of angle - based for rectangle - to - parallelogram proof) or misstate the logical flow.

Answer:

B. For quadrilateral \(ABCD\), if \(m\angle A = m\angle B = m\angle C = m\angle D = 90^\circ\), then \(\overline{AB} \parallel \overline{DC}\) and \(\overline{AD} \parallel \overline{BC}\)