Question

given: abcd is a parallelogram, \\(\overline{ac}\\) bisects \\(\angle bcd\\) and \\(\overline{db}\\) bisects \\(\angle abc\\). prove: \\(\overline{ac} \perp \overline{bd}\\). step statement reason 1 abcd is a parallelogram \\(\overline{ac}\\) bisects \\(\angle bcd\\) \\(\overline{db}\\) bisects \\(\angle abc\\) given 2 \\(\angle abc\\) and \\(\angle bcd\\) are supplementary consecutive angles in a parallelogram are supplementary 3 \\(\angle bca \cong \angle acd\\) an angle bisector divides an angle into two congruent angles 4 \\(\angle abd \cong \angle dbc\\) an angle bisector divides an angle into two congruent angles 5 \\(\overline{ad} \parallel \overline{bc}\\) opposite sides of a parallelogram are parallel 6 \\(\overline{ab} \parallel \overline{cd}\\) opposite sides of a parallelogram are parallel 7 \\(\angle abd \cong \angle bdc\\) parallel lines cut by a transversal form congruent alternate interior angles 8 \\(\angle bac \cong \angle acd\\) parallel lines cut by a transversal form congruent alternate interior angles 9 \\(\angle dbc \cong \angle bdc\\) transitive property 10 \\(\triangle bcd\\) is an isosceles triangle the triangle has two congruent angles 11 \\(\overline{ad} \cong \overline{bc}\\) opposite sides of a parallelogram are congruent 12 \\(\overline{ab} \cong \overline{cd}\\) opposite sides of a parallelogram are congruent 13 \\(m\angle dbc = \frac{1}{2}\angle abc\\) when an angle is bisected, each part is half the measure of the whole angle 14 \\(m\angle bca = \frac{1}{2}\angle bcd\\) when an angle is bisected, each part is half the measure of the whole angle try type of statement

Explanation:

Step1: Analyze angle sum in parallelogram

In parallelogram \(ABCD\), \(\angle ABC + \angle BCD = 180^\circ\) (consecutive angles supplementary). From angle bisectors, \(\angle DBC=\frac{1}{2}\angle ABC\), \(\angle BCA = \frac{1}{2}\angle BCD\). So \(\angle DBC+\angle BCA=\frac{1}{2}(\angle ABC + \angle BCD)=90^\circ\).

Step2: Find angle between diagonals

In \(\triangle BOC\) (where \(O\) is intersection of \(AC\) and \(BD\)), \(\angle BOC = 180^\circ-(\angle DBC + \angle BCA)=90^\circ\). Thus, \(AC\perp BD\).

Answer:

\(\overline{AC} \perp \overline{BD}\) is proven.