Question

given: e is the midpoint of \\(\overline{bd}\\) and \\(\overline{ac} \perp \overline{bd}\\).
prove: \\(\triangle bae \cong \triangle dae\\).
step | statement | reason
1 | e is the midpoint of \\(\overline{bd}\\) \\(\overline{ac} \perp \overline{bd}\\) | given
(try dropdown with type of statement options: \\(\overline{xy} \cong \overline{wz}\\), \\(\angle x \cong \angle y\\), \\(\triangle xyz \cong \triangle tuv\\), \\(\angle x\\) is a right angle, classify a triangle) and a diagram of quadrilateral with points a, b, c, d and e as intersection of diagonals, ac perpendicular to bd.

Explanation:

Step1: Analyze given info

E is midpoint of \( \overline{BD} \), so \( \overline{BE} \cong \overline{DE} \) (midpoint definition). \( \overline{AC} \perp \overline{BD} \), so \( \angle BEA \) and \( \angle DEA \) are right angles (perpendicular lines form right angles), thus \( \angle BEA \cong \angle DEA \).

Step2: Identify common side

\( \overline{AE} \cong \overline{AE} \) (reflexive property).

Step3: Apply SAS congruence

In \( \triangle BAE \) and \( \triangle DAE \): \( \overline{BE} \cong \overline{DE} \), \( \angle BEA \cong \angle DEA \), \( \overline{AE} \cong \overline{AE} \). By SAS (Side - Angle - Side) congruence criterion, \( \triangle BAE \cong \triangle DAE \). For the statement in the table (step - by - step proof continuation), since \( \overline{AC} \perp \overline{BD} \), \( \angle BEA \) and \( \angle DEA \) are right angles, so the statement " \( \angle BEA \) is a right angle" (or similar, but from the dropdown, " \( \angle X \) is a right angle" - here \( X \) can be \( BEA \) or \( DEA \)) fits. So the correct statement from the dropdown is " \( \angle X \) is a right angle" (as \( \overline{AC} \perp \overline{BD} \) implies right angles at E for those triangles).

Answer:

\( \angle X \) is a right angle (the option " \( \angle X \) is a right angle" from the dropdown, as it relates to the right angles formed by \( \overline{AC} \perp \overline{BD} \) which is key for proving triangle congruence via SAS)