Question

prove: \\(\overline{ac} \perp \overline{bd}\\).
step\tstatement\treason
1\t\\(abcd\\) is a parallelogram
\\(\overline{ac}\\) bisects \\(\angle bcd\\)
\\(\overline{db}\\) bisects \\(\angle abc\\)\tgiven
2\t\\(\angle bca \cong \angle acd\\)\tan angle bisector divides an angle into two congruent angles
3\t\\(\angle abd \cong \angle dbc\\)\tan angle bisector divides an angle into two congruent angles
4\t\\(\overline{ad} \cong \overline{bc}\\)\topposite sides of a parallelogram are congruent
5\t\\(\overline{ab} \parallel \overline{cd}\\)\topposite sides of a parallelogram are parallel
6\t\\(\overline{ab} \cong \overline{cd}\\)\topposite sides of a parallelogram are congruent
7\t\\(\overline{bc} \parallel \overline{ad}\\)\topposite sides of a parallelogram are parallel
8\t\\(\angle abc\\) and \\(\angle bcd\\) are supplementary\tconsecutive angles in a parallelogram are supplementary
9\t\\(m\angle bcd = 2\angle bca\\)\twhen an angle is bisected, the whole angle is twice the measure of each part
10\t\\(m\angle abc = 2\angle dbc\\)\twhen an angle is bisected, the whole angle is twice the measure of each part
try type of statement

Explanation:

Step1: Analyze consecutive angles

From step 8, \( \angle ABC + \angle BCD = 180^\circ \) (supplementary). From steps 9 and 10, \( m\angle BCD = 2\angle BCA \), \( m\angle ABC = 2\angle DBC \). Substitute: \( 2\angle DBC + 2\angle BCA = 180^\circ \), simplify to \( \angle DBC + \angle BCA = 90^\circ \).

Step2: Find angle at intersection

In \( \triangle BEC \), \( \angle BEC = 180^\circ - (\angle DBC + \angle BCA) = 180^\circ - 90^\circ = 90^\circ \). So \( \overline{AC} \perp \overline{BD} \) (definition of perpendicular, \( 90^\circ \) angle between lines).

Answer:

By showing \( \angle BEC = 90^\circ \) (using supplementary angles and angle bisector properties), we prove \( \overline{AC} \perp \overline{BD} \).