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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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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