Question

how can it be proven that \\( \overline { a c } \cong \overline { d b } \\) ? drag tiles to the empty boxes to complete the proof correctly.

\\( a b c d \\) is a rectangle because that is given.

\\( \overline { a d } \cong \square \\) because of the reflexive property.

\\( \overline { a b } \cong \square \\) because opposite sides of rectangles are congruent.

\\( m \angle a d c = m \angle b a d \\) because all angles in a rectangle are right angles and thus, \\( \angle a d c \cong \angle b a \\)

\\( \delta a c d \cong \delta d b a \\) because of \\( \square \\).

\\( \overline { a c } \cong \overline { d b } \\) because corresponding parts of congruent triangles are congruent.

\\( \overline { a d } \quad \overline { a b } \quad \overline { b c } \quad \overline { d c } \quad a s a \quad s a s \\)

section 1: item 9 of 12

Explanation:

Step1: Reflexive property

By the reflexive property of congruence, any segment is congruent to itself. So, $\overline{AD}\cong\overline{AD}$.

Step2: Opposite sides of rectangle

In a rectangle \(ABCD\), opposite sides are congruent. So, \(\overline{AB}\cong\overline{DC}\).

Step3: Triangle congruence criterion

We have two sides and the included angle. \(\overline{AD}\cong\overline{AD}\) (reflexive), \(\overline{AB}\cong\overline{DC}\) (opposite sides of rectangle), \(\angle ADC\cong\angle BAD\) (right angles). So, \(\triangle ACD\cong\triangle DBA\) by the SAS (Side - Angle - Side) congruence criterion.

Answer:

\(\overline{AD}\cong\overline{AD}\), \(\overline{AB}\cong\overline{DC}\), \(\triangle ACD\cong\triangle DBA\) because of \(SAS\)