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
| \\(\angle \square\\) is a right angle | perpendicular lines form right angles

Explanation:

Step1: Analyze Perpendicular Lines

Since \( \overline{AC} \perp \overline{BD} \), the angles formed at their intersection (point \( E \)) are right angles. So \( \angle AEB \) (or \( \angle AED \)) is a right angle. Looking at the triangles \( \triangle BAE \) and \( \triangle DAE \), the angle at \( E \) between \( AC \) and \( BD \) for these triangles is \( \angle AEB \) and \( \angle AED \), but in the context of the proof for \( \triangle BAE \cong \triangle DAE \), the right angle from \( AC \perp BD \) is \( \angle AEB \) (or \( \angle AED \); here, for \( \triangle BAE \), the right angle is \( \angle AEB \)). Wait, actually, when \( AC \perp BD \), the angles \( \angle AEB \) and \( \angle AED \) are right angles. So the angle that is a right angle here, for the step, is \( \angle AEB \) (or \( \angle AED \); but let's check the diagram. The diagram shows \( E \) as the intersection of \( AC \) and \( BD \). So \( AC \perp BD \) means \( \angle AEB = 90^\circ \) and \( \angle AED = 90^\circ \). So the angle in the box should be \( AEB \) (or \( AED \); but since we are proving \( \triangle BAE \cong \triangle DAE \), let's see: \( \triangle BAE \) has angle at \( E \): \( \angle AEB \), and \( \triangle DAE \) has angle at \( E \): \( \angle AED \). But the reason is "Perpendicular lines form right angles", so the angle formed by \( AC \) and \( BD \) at \( E \) is a right angle. So the angle is \( \angle AEB \) (or \( \angle AED \)). Let's confirm: \( AC \perp BD \), so the intersection at \( E \) creates right angles. So the angle is \( \angle AEB \) (or \( \angle AED \)). So the answer is \( \angle AEB \) (or \( \angle AED \); but likely \( \angle AEB \) as we are looking at \( \triangle BAE \)).

Step2: Confirm the Angle

Since \( AC \) is perpendicular to \( BD \) at \( E \), the angle \( \angle AEB \) (and \( \angle AED \)) is a right angle. So the angle in the box is \( \angle AEB \) (or \( \angle AED \); but for \( \triangle BAE \), it's \( \angle AEB \)).

Answer:

\( \angle AEB \) (or \( \angle AED \); the correct angle here is \( \angle AEB \) or \( \angle AED \), but based on the triangle \( \triangle BAE \), it's \( \angle AEB \))