Diagonals of Parallelogram Bisect Each Other

In this post, we show that the diagonals of a parallelogram bisect each other. We will use the fact that the diagonal of a parallelogram is a transversal to the opposite sides of the parallelogram. The transversal and the sides form congruent alternate interior angles. Recall that the alternate interior angles formed by a transversal and parallel lines are congruent.

Given

Parallelogram $ABCD$ with diagonals $\overline{AC}$ and $\overline{BD}$ .

What to Show

$\overline{AO} \cong \overline{OC}$ and $\overline{BO} \cong \overline{OD}$ .

Proof

$\angle OAB \cong \angle OCD$ because they are alternate interior angles of parallel lines that pass through $\overline {AB}$ and $\overline {CD}$ cut by the transversal $\overline{AC}$ (A).

$\overline{AB} \cong \overline{CD}$ because opposite sides of a parallelogram are congruent (S).

$\angle ABO \cong \angle ODC$ because they are alternate interior angles of parallel lines that pass through $\overline {AB}$ and $\overline {CD}$ cut by the transversal $\overline{BD}$ (A).

By ASA Triangle Congruence theorem, $\triangle AOB \cong \triangle COD$ .

$\overline{AO} \cong \overline{OC}$ and $\overline{BO} \cong \overline{OD}$ because corresponding sides of congruent triangles are congruent.

Given

What to Show

Proof

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