This is the first of a series of post about the conditions of a quadrilateral to be a parallelogram. In this post, I will be discussing the proof that if the opposite sides of a quadrilateral are congruent, then it is a parallelogram.

**Given**

Quadrilateral $latex ABCD$

$latex \overline{AB} \cong \overline{CD}$

$latex \overline {BC} \cong \overline{AD}$

**Proof**

Draw diagonal $latex \overline{BD}$.

Now, $latex \overline{AB} \cong \overline{CD}$ (S) and $latex \overline {BC} \cong \overline{AD}$ (S).

Also, a segment is congruent to itself (Relfexive Property), so $latex \overline{BD} \cong \overline{BD}$ (S).

Therefore, by the SSS Triangle Congruence, $latex \triangle ABD \cong \triangle BDA$.

Since corresponding angles of congruent triangles are congruent,

$latex \angle ADB \cong \angle CBD$ and $latex \angle ABD \cong \angle CDB$.

But these pairs of angles are corresponding angles, so by the **Parallel Line Postulate**,

$latex AB \parallel DC$ and $latex AD \parallel BD$.

So, quadrilateral $latex ABCD$ is a parallelogram.

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