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11 ∠a ≅ ∠c corresponding parts of congruent triangles are congruent (cp…

Question

11 ∠a ≅ ∠c corresponding parts of congruent triangles are congruent (cpctc)
12 ∠b ≅ ∠d corresponding parts of congruent triangles are congruent (cpctc)
13 \\(\overline{ce} \cong \overline{ce}\\) reflexive property
14 \\(\overline{ae} \cong \overline{ae}\\) reflexive property
15 ∠eab ≅ ∠eab reflexive property
\\(\overline{bd} \cong \overline{ae}\\) × reflexive property
notice
to use the reflexive property you must state that a segment is
congruent to itself.
ok

Explanation:

Step1: Recall Reflexive Property

The reflexive property of congruence for segments states that a segment is congruent to itself, i.e., $\overline{XY} \cong \overline{XY}$ for any segment $\overline{XY}$.

Step2: Analyze the Incorrect Statement

In the given problem, the statement $\overline{BD} \cong \overline{AE}$ was incorrectly justified with the reflexive property. To use the reflexive property, we need to have a segment congruent to itself. So we should replace either $\overline{BD}$ or $\overline{AE}$ with itself. For example, if we consider $\overline{BD}$, the correct congruence statement using the reflexive property is $\overline{BD} \cong \overline{BD}$. Similarly, for $\overline{AE}$, it would be $\overline{AE} \cong \overline{AE}$. Since the original incorrect statement involved $\overline{BD}$ and $\overline{AE}$, the correct way to apply the reflexive property is to have a segment congruent to itself, so a correct statement would be $\overline{BD} \cong \overline{BD}$ (or $\overline{AE} \cong \overline{AE}$).

Answer:

$\overline{BD} \cong \overline{BD}$ (or $\overline{AE} \cong \overline{AE}$ but since the incorrect one was $\overline{BD} \cong \overline{AE}$, the correct reflexive should be a segment congruent to itself, so $\overline{BD} \cong \overline{BD}$ or other self - congruent segment, but for the given incorrect pair $\overline{BD} \cong \overline{AE}$, the correct reflexive application would be a segment congruent to itself, e.g., $\overline{BD} \cong \overline{BD}$)