Question

given: c is midpoint of \\( \overline{bd} \\)
\\( \overline{ab} \perp \overline{bd} \\)
\\( \overline{bd} \perp \overline{de} \\)
prove: \\( \triangle abc \cong \triangle edc \\)

1. c is midpoint of \\( \overline{bd} \\)
2. \\( \overline{ab} \perp \overline{bd} \\) and \\( \overline{bd} \perp \overline{de} \\)
3. \\( \overline{bc} \cong \overline{cd} \\)
4. \\( \angle bca \cong \angle ecd \\)
5. \\( \angle abc \\) and \\( \angle edc \\) are right angles
6. \\( \angle abc \cong \angle edc \\)
7. \\( \triangle abc \cong \triangle edc \\)

Explanation:

Step1: Given

Given in the problem statement.

Step2: Given

Given in the problem statement.

Step3: Definition of midpoint

If \(C\) is the mid - point of \(\overline{BD}\), then \(BC = CD\), so \(\overline{BC}\cong\overline{CD}\) (by the definition of congruent segments).

Step4: Vertical angles

\(\angle BCA\) and \(\angle ECD\) are vertical angles. Vertical angles are congruent, so \(\angle BCA\cong\angle ECD\).

Step5: Definition of perpendicular lines

Since \(\overline{AB}\perp\overline{BD}\), \(\angle ABC = 90^{\circ}\). Since \(\overline{BD}\perp\overline{DE}\), \(\angle EDC=90^{\circ}\). So \(\angle ABC\) and \(\angle EDC\) are right angles.

Step6: All right angles are congruent

Since \(\angle ABC = 90^{\circ}\) and \(\angle EDC = 90^{\circ}\), \(\angle ABC\cong\angle EDC\) (by the property that all right angles are congruent).

Step7: ASA (Angle - Side - Angle)

In \(\triangle ABC\) and \(\triangle EDC\), we have \(\angle ABC\cong\angle EDC\), \(\overline{BC}\cong\overline{CD}\), \(\angle BCA\cong\angle ECD\). By the ASA (Angle - Side - Angle) congruence criterion, \(\triangle ABC\cong\triangle EDC\).

Answer:

1. Given
2. Given
3. Definition of mid - point
4. Vertical angles are congruent
5. Definition of perpendicular lines
6. All right angles are congruent
7. ASA (Angle - Side - Angle) congruence criterion