Question

1. given: c is the midpoint of ab and \\(\angle acd \cong \angle bcd\\)

prove: dc is a perpendicular bisector of ab

Explanation:

Identify given information

We are given:

1. \(C\) is the midpoint of segment \(AB\).
2. \(\angle ACD \cong \angle BCD\).

Use midpoint definition

Since \(C\) is the midpoint of \(AB\), by definition:

$$AC = BC$$

Thus, \(DC\) bisects \(AB\).

Apply linear pair postulate

Points \(A\), \(C\), and \(B\) lie on a straight line.
Therefore, \(\angle ACD\) and \(\angle BCD\) form a linear pair:

$$m\angle ACD + m\angle BCD = 180^\circ$$

Solve for angle measures

Since \(\angle ACD \cong \angle BCD\), their measures are equal:

$$m\angle ACD = m\angle BCD$$

Substituting this into the linear pair equation:

$$2 \cdot m\angle ACD = 180^\circ \implies m\angle ACD = 90^\circ$$

Thus, \(DC \perp AB\).

Conclude the proof

Since \(DC\) is perpendicular to \(AB\) and bisects \(AB\) at \(C\):
\(DC\) is the perpendicular bisector of \(AB\).

Answer:

Since \(C\) is the midpoint of \(AB\), we have \(AC = BC\), which means \(DC\) bisects \(AB\).

Because \(A\), \(C\), and \(B\) form a straight line, \(\angle ACD\) and \(\angle BCD\) are a linear pair, meaning:

$$m\angle ACD + m\angle BCD = 180^\circ$$

Given that \(\angle ACD \cong \angle BCD\), their measures are equal:

$$m\angle ACD = m\angle BCD = 90^\circ$$

This proves that \(DC \perp AB\). Since \(DC\) is both perpendicular to \(AB\) and bisects \(AB\), \(DC\) is the perpendicular bisector of \(AB\).