Theorems: Conditions for a Quadrilateral to be a Parallelogram

Theorem 1: A quadrilateral is a parallelogram if its opposite sides are equal.

Given: A quadrilateral ABCD in which AB = CD and BC = DA.
To Prove: ABCD is a parallelogram.
Construction: Join AC.

Proof: In $\Delta ACBand\Delta CAD$
AC = CA [Common]
CB = AD [Given]
AB = CD [Given]
So, by SSS congruence Criterion,  $\Delta ACB\cong \Delta CAD$

$\begin{align}
& \angle CAB=\angle ACD  [CPCT] \\
& \angle ACB=\angle CAD  [CPCT] \\
\end{align}$
$ \angle CAB=\angle ACD$   and they are alternate interior angles
∴ $AB\parallel DC...(i)$

$\angle ACB=\angle CAD$   and they are alternate interior angles
∴ $BC\parallel AD...(ii)$

From (i) and (ii),
$AB\parallel DCandBC\parallel AD.$

Hence, ABCD is a parallelogram.



Theorem 2: A quadrilateral is a parallelogram if its opposite angles are equal.

Given: A Quadrilateral ABCD in which $\angle A=\angle Cand\angle B=\angle D.$
To Prove: ABCD is a parallelogram.

Proof: In quad. ABCD
$\begin{align}
& \angle A=\angle C \\
& \angle B=\angle D \\
& \angle A+\angle B=\angle C+\angle D...(i) \\
\end{align}$

Since sum of the angles of a quadrilateral is $360{}^\circ $.
$\begin{align}
& \angle A+\angle B+\angle C+\angle D=360{}^\circ \\

& (\angle A+\angle B)+(\angle A+\angle B)=360{}^\circ \\

& 2(\angle A+\angle B)=360{}^\circ \\

& \angle A+\angle B=180{}^\circ \\

& \angle A+\angle B=\angle C+\angle D=180{}^\circ from(i) \\
\end{align}$

AB intersects AD and BC at A and B and $\angle A+\angle B=180{}^\circ $ 
 i.e, sum of consecutive interior angles is $180{}^\circ $.
$AD\parallel BC...(ii)$

Again,
$\begin{align}
& \angle A+\angle B=180{}^\circ \\

& \Rightarrow \angle C+\angle B=180{}^\circ [∵\angle A=\angle C] \\

\end{align}$

BC intersects AB and DC at A and C and $\angle B+\angle C=180{}^\circ $  
i.e, sum of consecutive interior angle is $180{}^\circ $.
$ AB\parallel DC...(iii)$

From (ii) and (iii), we get
$AD\parallel BCandAB\parallel DC.$
Hence, ABCD is a parallelogram.



Theorem 3: If the diagonals of a quadrilateral bisect each other then the quadrilateral is a parallelogram.

Given: A quadrilateral ABCD in which the diagonals AC and BD intersect at O such that AO = OC and BO = OD.
To Prove: Quad. ABCD is a parallelogram.
Proof: In $\Delta AODand\Delta COB$
AO = OC   [Given]
OD = OB   [Given]
$\angle AOD=\angle COB [VOA]$
So, by SAS congruence criterion,
$\begin{align}
& \Delta AOD\cong COB \\

& \angle OAD=\angle OCB[CPCT] \\
\end{align}$

$\angle OAD=\angle OCB$   and they are alternate interior angles
∴ $AD\parallel BC$

Similarly, ∴$AB\parallel CD$
Hence, ABCD is a parallelogram.



Theorem 4: A quadrilateral is a parallelogram, if its one pair of opposite sides are equal and parallel.

Given: A quad. ABCD in which AB = CD and AB || CD.
To Prove: ABCD is a parallelogram.
Construction: Join AC.

Proof: In $\Delta ABCand\Delta CDA$
AB = DC [Given]
AC = AC [Common]
$\angle BAC=\angle DCA  [∵ AB\parallel CD,Alt.Int.angle]$
So, SAS congruence criterion ,
$\begin{align}
& \Delta ABC\cong \Delta CDA \\

& \Rightarrow \angle BCA=\angle DAC[CPCT] \\
\end{align}$

$\angle BCA=\angle DAC$   and they are alternate interior angles
∴ AD || BC.

Thus, AB || CD and AD || BC.
Hence, quadrilateral ABCD is a parallelogram.


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