given: e is the midpoint of $\overline{bd}$ and $\overline{ac} \perp \overline{bd}$. prove: $\triangle bae \cong \triangle dae$. step 1 statement: e is the midpoint of $\overline{bd}$, $\overline{ac} \perp \overline{bd}$ reason: given. $\angle \square$ is a right angle reason: perpendicular lines form right... (with a diagram of a quadrilateral with points b, c, e, etc.)

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# Explanation:
## Step1: Analyze Perpendicular Lines
Since $ \overline{AC} \perp \overline{BD} $, the angles formed at their intersection (point $ E $) are right angles. So, $ \angle BEA $ and $ \angle DEA $ are right angles.
## Step2: Identify the Angle
For the triangles $ \triangle BAE $ and $ \triangle DAE $, the right angles are $ \angle BEA $ (or $ \angle DEA $, but in the context of the proof step, we look at the angle related to the sides. Since $ E $ is on $ BD $ and $ AC \perp BD $, the angle at $ E $ between $ AC $ and $ BD $ is the right angle. So the angle is $ \angle BEA $ (or $ \angle AEB $, $ \angle AED $, $ \angle DEA $). But from the diagram and the triangles $ \triangle BAE $ and $ \triangle DAE $, the right angles are $ \angle BEA $ and $ \angle DEA $. So the angle in the box should be $ BEA $ (or $ AEB $, $ AED $, $ DEA $). Let's confirm: $ AC \perp BD $ implies $ \angle AEB = 90^\circ $ and $ \angle AED = 90^\circ $. For $ \triangle BAE $ and $ \triangle DAE $, the right angles are $ \angle BEA $ and $ \angle DEA $ respectively. So the angle is $ BEA $ (or $ AEB $).

# Answer:
The angle is $ \angle BEA $ (or $ \angle AEB $, $ \angle AED $, $ \angle DEA $; most appropriately $ \angle BEA $ or $ \angle AEB $ for $ \triangle BAE $ and $ \triangle DAE $ context). So the box should be filled with $ BEA $ (or $ AEB $, etc.).

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QUESTION IMAGE

Question

Explanation:

Step1: Analyze Perpendicular Lines

Step2: Identify the Angle

Answer: