Quadrilaterals with Congruent Opposite Angles are Parallelograms

In last week’s post, we have learned that quadrilaterals with congruent opposite sides are parallelograms. In this post, we show a related theorem. That is, quadrilaterals whose opposite angles are congruent are parallelograms. In the figure above, $ABCD$ is a quadrilateral where $\angle A$ and $\angle C$ are congruent and $\angle D$ and $\angle B$ are congruent. In proving the theorem, we need to show that the opposite sides of $ABCD$ are parallel. That is $\overline {AB}$ is parallel to $\overline{CD}$ and $\overline {AD}$ is parallel to $\overline {BC}$.

Theorem

If the opposite angles of a quadrilateral are congruent, then the quadrilateral is a parallelogram.

Given $ABCD$ with $A \cong C$ and $B \cong D$.

What To Show $AB \parallel CD$ and $AD \parallel BC$.

Proof

Since both $ABC$ and $ADC$ are triangles, by the Triangle Angle Sum Theorem, the sum of their interior angles are equal (both $180^\circ$).

Therefore, $m \angle 1 + m \angle 2 + m \angle D = m \angle 3 + m \angle 4 + m \angle B$.

Since from the given, $m \angle B \cong m \angle D$, $m \angle 1 + m \angle 2 = m \angle 3 + m \angle 4.$ (1)

Also, since $\angle A \cong \angle C$, $m \angle 1 + m \angle 3 = m \angle 2 + m \angle 4.$ (2)

Subtracting (2) from (1), we have $m \angle 2 - m \angle 3 = m \angle 3 - m \angle 2$

which results to $m \angle 2 = m \angle 3$.

Therefore $\angle 2 \cong \angle 3$.

Since $\angle 2$ and $\angle 3$ are congruent, by the Parallel Postulate, $AD \parallel BC$.

Using the steps similar to the proof above, it can be shown that $\overline{AD} \parallel \overline{BC}$ (left as an exercise).

Therefore, $ABCD$ is a parallelogram.

So, if opposite angles of a quadrilateral are congruent, then the quadrilateral is a parallelogram.