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, $latex ABCD$ is a quadrilateral where $latex \angle A$ and $latex \angle C$ are congruent and $latex \angle D$ and $latex \angle B$ are congruent. In proving the theorem, we need to show that the opposite sides of $latex ABCD$ are parallel. That is $latex \overline {AB}$ is parallel to $latex \overline{CD}$ and $latex \overline {AD}$ is parallel to $latex \overline {BC}$.

**Theorem**

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

**Given **

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

**What To Show**

$latex AB \parallel CD$ and $latex AD \parallel BC$.

**Proof**

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

Therefore,

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

Since from the given, $latex m \angle B \cong m \angle D$,

$latex m \angle 1 + m \angle 2 = m \angle 3 + m \angle 4.$ (1)

Also, since $latex \angle A \cong \angle C$,

$latex m \angle 1 + m \angle 3 = m \angle 2 + m \angle 4.$ (2)

Subtracting (2) from (1), we have

$latex m \angle 2 – m \angle 3 = m \angle 3 – m \angle 2$

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

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

Since $latex \angle 2$ and $latex \angle 3$ are congruent, by the Parallel Postulate,

$latex AD \parallel BC$.

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

Therefore, $latex ABCD$ is a parallelogram.

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