Question

below is a proof that the diagonals of a parallelogram bisect each other. which of the following best completes the missing reason?drawstepstatementreason1$abcd$ is a parallelogram with diagonals $overline{ac}$ and $overline{bd}$given2$angle 1 cong angle 4$ and $angle 2 cong angle 3$alternate interior angle theorem$\bigcirc$ opposite sides of a parallelogram are congruent$\bigcirc$ vertical angles are always congruent$\bigcirc$ definition of midpoint$\bigcirc$ $overline{bd}$ bisects the opposite angles

Explanation:

In parallelogram \(ABCD\), \(AB \parallel CD\) and \(AD \parallel BC\). When transversal \(AC\) cuts \(AB\) and \(CD\), \(\angle 1\) and \(\angle 2\) are alternate interior angles; when transversal \(BD\) cuts \(AB\) and \(CD\), \(\angle 3\) and \(\angle 4\) are alternate interior angles. The Alternate Interior Angle Theorem states that alternate interior angles formed by a transversal cutting two parallel lines are congruent, which directly justifies \(\angle 1 \cong \angle 4\) and \(\angle 2 \cong \angle 3\).

Answer:

Alternate Interior Angle Theorem