Question

given: c is the midpoint of \\(\overline{bd}\\). prove: \\(\triangle acb \cong \triangle acd\\) complete the two - column proof. diagram: triangle with vertices a (top), b (left - bottom), d (right - bottom), c (mid - bottom of bd), right angles at c between ac and bc, ac and dc statements 1. c is the midpoint of \\(\overline{bd}\\) 1. given 2. \\(\overline{bc} \cong \overline{cd}\\) 2. ♦ 3. \\(\overline{ac} \cong \overline{ac}\\) 3. reflexive property 4. \\(\angle bca\\) and \\(\angle dca\\) are right \\(\angle s\\) 4. given 5. ♦ 5. all right \\(\angle s\\) are \\(\cong\\) 6. \\(\triangle acb \cong \triangle acd\\) 6. sas options: angles bca and dca form a linear pair, angle bca is congruent to angle dca, segment ac is perpendicular to segment bd

Explanation:

Step1: Find reason for \( \overline{BC} \cong \overline{CD} \)

The definition of a midpoint states that a midpoint of a segment divides it into two congruent segments. Since \( C \) is the midpoint of \( \overline{BD} \), by the definition of midpoint, \( \overline{BC} \cong \overline{CD} \).

Step2: Find statement for step 5

The reason for step 5 is "all right \( \angle s \) are \( \cong \)", so the statement should be that the two right angles ( \( \angle BCA \) and \( \angle DCA \)) are congruent. So \( \angle BCA \cong \angle DCA \).

Answer:

For step 2 reason: definition of midpoint
For step 5 statement: \( \angle BCA \cong \angle DCA \)