Question

given: d is the midpoint of ac, ba ≅ bc and ∠eda ≅ ∠fdc. prove: △aed ≅ △cfd. note: quadrilateral properties are not permitted in this proof. step statement reason 1 d is the midpoint of \\(\overline{ac}\\) \\(\overline{ba} ≅ \overline{bc}\\) ∠eda ≅ ∠fdc given try type of statement

Explanation:

Step1: Identify Midpoint Property

Since \( D \) is the midpoint of \( \overline{AC} \), by the definition of a midpoint, \( \overline{AD} \cong \overline{CD} \).

Step2: Identify Isosceles Triangle Angle

Given \( \overline{BA} \cong \overline{BC} \), \( \triangle BAC \) is isosceles with \( \angle A \cong \angle C \) (base angles of an isosceles triangle are congruent).

Step3: Apply ASA Congruence

We have \( \angle EDA \cong \angle FDC \) (given), \( \overline{AD} \cong \overline{CD} \) (from midpoint), and \( \angle A \cong \angle C \) (from isosceles triangle). By the Angle - Side - Angle (ASA) congruence criterion, \( \triangle AED \cong \triangle CFD \).

Answer:

To prove \( \triangle AED \cong \triangle CFD \):

1. \( \overline{AD} \cong \overline{CD} \) (Definition of midpoint, as \( D \) is the midpoint of \( \overline{AC} \)).
2. \( \angle A \cong \angle C \) (Base angles of isosceles triangle \( \triangle BAC \) since \( \overline{BA} \cong \overline{BC} \)).
3. \( \angle EDA \cong \angle FDC \) (Given).

By ASA (Angle - Side - Angle) congruence postulate, \( \triangle AED \cong \triangle CFD \).