Theorem 1: A diagonal of a parallelogram divides it into two congruent triangles.
Given: A parallelogram ABCD
To Prove: A diagonal AC of parallelogram ABCD divides it into congruent triangles ABC and CDA i.e $\Delta ABC\cong \Delta CDA$ .
Construction: Join AC.
Proof: Since ABCD is a parallelogram,
$ AB\parallel DC$ and $AD\parallel BC$ and AC intersect at A and C respectively.
So, $\angle DAC=\angle BCA...(i)$ [Alternate interior angles]
$AB\parallel DC$ and transversal AC intersect them at A and C respectively.
$\angle BAC=\angle DCA...(ii)$ [Alternate interior angles]
Now, In $\Delta ABCand\Delta CDA$
$\begin{align}
& \angle BCA=\angle DAC [from(i)] \\
& AC=AC [Common] \\
& \angle BAC=\angle DCA [from(ii)] \\
\end{align}$
$\Delta ABC\cong \Delta CDA$
By ASA congruence criteria.
Theorem 2: In a parallelogram, opposite sides are equal.
Given: A parallelogram ABCD
To Prove: AB = CD and DA = BC
Construction: Join AC.
Proof: Since ABCD ia a parallelogram,
∴ $AD\parallel BC$ and AC intersect them at A and C respectively.
So, $\angle DAC=\angle BCA...(i)$ [Alternate interior angle]
$AB\parallel DC$ and AC intersect them at A and C respectively.
$\angle BAC=\angle DCA...(ii)$
In $\Delta ADCand\Delta CBA$
$\begin{align}
& \angle DAC=\angle BCA [from(i)] \\
& AC=AC [Common] \\
& \angle DCA=\angle BAC [from(ii)] \\
\end{align}$
$\Delta ADC\cong \Delta CBA$ By ASA congruence criteria
AD = CB and DC = BA [CPCT]
Theorem 3: The opposite angles of a parallelogram are equal.
Given: A Parallelogram ABCD
To Prove: $\angle A=\angle C$ and $\angle B=\angle D$
Proof: Since ABCD is a parallelogram,
AB || DC and AD is transversal
∴ $\angle A+\angle D=180{}^\circ ...(i)$ [Sum of consecutive interior angles is 180]
AD || BC and DC is a transversal
∴ $\angle D+\angle C=180{}^\circ ...(ii)$ &êmsp; [sum of consecutive interior angles is 180]
From (i) and (ii)
$\begin{align}
& \angle A+\angle D=\angle D+\angle C \\
& \angle A=\angle C \\
\end{align}$
Similarly, $\angle B=\angle D.$
Hence, $\angle A=\angle Cand\angle B=\angle D$
Theorem 4: The diagonals of a parallelogram bisect each other.
Given: A Parallelogram ABCD such that its diagonals AC and BD intersect at O.
To Prove: OA = OC and OB = OD
Proof: Since ABCD is a parallelogram,
AB || DC and AC intersects them at A and C.
∴ $\angle BAC=\angle DCA$ [Alternate interior angles are equal]
$\angle BAO=\angle DCO$
AB || DC and BD intersects them at B and D.
∴ $\angle ABD=\angle CDB$ [Alternate interior angles are equal]
$\angle ABO=\angle CDO$
In $\Delta AOBand\Delta COD$
$\begin{align}
& \angle BAO=\angle DCO [from(i)] \\
& AB=CD [opposite sides ||gm] \\
& \angle ABO=\angle CDO [from(ii)] \\
\end{align}$
$\Delta AOB\cong \Delta COD$
By ASA congruence criterion
OA = OC and OB = OD [CPCT]
Hence, OA = OC and OB = OD.
Theorem 5: In a parallelogram, the bisectors of any two consecutive angles intersect at right angle.
Given: A parallelogram ABCD such that the bisectors of consecutive angles A and B intersect at P.
To Prove: $\angle APB=90{}^\circ $
Proof: Since ABCD is a parallelogram,
AD || BC and transversal AB intersect them.
∴ $\angle A+\angle B=180{}^\circ $ [sum of consecutive interior angles is 180]
$\begin{align}
& \frac{1}{2}\angle A+\frac{1}{2}\angle B=90{}^\circ \\
& \angle 1+\angle 2=90{}^\circ ...(i) \\
\end{align}$
In $\Delta APB$
$\angle 1+\angle APB+\angle 2=180{}^\circ $
$\begin{align}
& 90{}^\circ +\angle APB=180{}^\circ [from(i)] \\
& \angle APB=90{}^\circ \\
\end{align}$
Theorem 6 : If a diagonal of a parallelogram bisects one of the angles of the parallelogram, it also bisects the second angle. Also, prove that it is a rhombus.
Given: A parallelogram ABCD in which diagonal AC bisects $\angle A$ .
To Prove: AC bisects $\angle C$ and ABCD is a rhombus.
Proof: Since ABCD is a parallelogram,
AB || DC and AC intersects them
∴ $\angle 1=\angle 3...(i)$ [Alternate interior angles]
AD || BC and AC intersects them
∴ $\angle 2=\angle 4...(ii)$ [Alternate interior angles]
But AC is bisector of $\angle C$ ,
∴ $\angle 1=\angle 2...(iii)$
From (i), (ii), (iii) we get
$\angle 3=\angle 4...(iv)$
Hence, AC bisects $\angle C$
From (ii) and (iii), we have
$\angle 1=\angle 4$
BC = AB [Angles opposite to equal sides are equal]
But, AB = DC and BC= AD [ABCD is a parallelogram]
∴ AB = BC = CD = DA.
Hence, ABCD is a rhombus.
Theorem 7: The angle bisectors of a parallelogram form a rectangle.
Given: A parallelogram ABCD in which bisectors of angles A, B, C, D intersect at P, Q, R, S to form a quadrilateral PQRS.
To Prove: PQRS is a rectangle.
Proof: Since ABCD is a parallelogram,
AD || BC and transversal AB intersect them at A and B respectively
∴ $\angle A+\angle B=180{}^\circ $ [Sum of consecutive interior angles is 180].
$\begin{align}
& \frac{1}{2}\angle A+\frac{1}{2}\angle B=90{}^\circ \\
& \angle BAS+\angle ABS=90{}^\circ ...(i) \\
\end{align}$
[AS and BS are bisectors of ㄥA and ㄥB]
In $\Delta ABS$ ,
$\angle BAS+\angle ABS+\angle ASB=180{}^\circ $ [sum of the angles of a triangle is 180]
$\begin{align}
& 90{}^\circ +\angle ASB=180{}^\circ \\
& \angle ASB=90{}^\circ \\
& \angle RSP=90{}^\circ [ㄥASB and ㄥRSP are vertically opposite angles, ㄥRSP = ㄥASB ] \\
\end{align}$
Similarly, $\angle SRQ=90{}^\circ ,\angle RQP=90{}^\circ and\angle SPQ=90{}^\circ $
Hence, PQRS is a rectangle.
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To Prove: A diagonal AC of parallelogram ABCD divides it into congruent triangles ABC and CDA i.e $\Delta ABC\cong \Delta CDA$ .
Construction: Join AC.
Proof: Since ABCD is a parallelogram,
$ AB\parallel DC$ and $AD\parallel BC$ and AC intersect at A and C respectively.
So, $\angle DAC=\angle BCA...(i)$ [Alternate interior angles]
$AB\parallel DC$ and transversal AC intersect them at A and C respectively.
$\angle BAC=\angle DCA...(ii)$ [Alternate interior angles]
Now, In $\Delta ABCand\Delta CDA$
$\begin{align}
& \angle BCA=\angle DAC [from(i)] \\
& AC=AC [Common] \\
& \angle BAC=\angle DCA [from(ii)] \\
\end{align}$
$\Delta ABC\cong \Delta CDA$
By ASA congruence criteria.
Theorem 2: In a parallelogram, opposite sides are equal.
To Prove: AB = CD and DA = BC
Construction: Join AC.
Proof: Since ABCD ia a parallelogram,
∴ $AD\parallel BC$ and AC intersect them at A and C respectively.
So, $\angle DAC=\angle BCA...(i)$ [Alternate interior angle]
$AB\parallel DC$ and AC intersect them at A and C respectively.
$\angle BAC=\angle DCA...(ii)$
In $\Delta ADCand\Delta CBA$
$\begin{align}
& \angle DAC=\angle BCA [from(i)] \\
& AC=AC [Common] \\
& \angle DCA=\angle BAC [from(ii)] \\
\end{align}$
$\Delta ADC\cong \Delta CBA$ By ASA congruence criteria
AD = CB and DC = BA [CPCT]
Theorem 3: The opposite angles of a parallelogram are equal.
To Prove: $\angle A=\angle C$ and $\angle B=\angle D$
Proof: Since ABCD is a parallelogram,
AB || DC and AD is transversal
∴ $\angle A+\angle D=180{}^\circ ...(i)$ [Sum of consecutive interior angles is 180]
AD || BC and DC is a transversal
∴ $\angle D+\angle C=180{}^\circ ...(ii)$ &êmsp; [sum of consecutive interior angles is 180]
From (i) and (ii)
$\begin{align}
& \angle A+\angle D=\angle D+\angle C \\
& \angle A=\angle C \\
\end{align}$
Similarly, $\angle B=\angle D.$
Hence, $\angle A=\angle Cand\angle B=\angle D$
Theorem 4: The diagonals of a parallelogram bisect each other.
Given: A Parallelogram ABCD such that its diagonals AC and BD intersect at O.
To Prove: OA = OC and OB = OD
Proof: Since ABCD is a parallelogram,
AB || DC and AC intersects them at A and C.
∴ $\angle BAC=\angle DCA$ [Alternate interior angles are equal]
$\angle BAO=\angle DCO$
AB || DC and BD intersects them at B and D.
∴ $\angle ABD=\angle CDB$ [Alternate interior angles are equal]
$\angle ABO=\angle CDO$
In $\Delta AOBand\Delta COD$
$\begin{align}
& \angle BAO=\angle DCO [from(i)] \\
& AB=CD [opposite sides ||gm] \\
& \angle ABO=\angle CDO [from(ii)] \\
\end{align}$
$\Delta AOB\cong \Delta COD$
By ASA congruence criterion
OA = OC and OB = OD [CPCT]
Hence, OA = OC and OB = OD.
Theorem 5: In a parallelogram, the bisectors of any two consecutive angles intersect at right angle.
To Prove: $\angle APB=90{}^\circ $
Proof: Since ABCD is a parallelogram,
AD || BC and transversal AB intersect them.
∴ $\angle A+\angle B=180{}^\circ $ [sum of consecutive interior angles is 180]
$\begin{align}
& \frac{1}{2}\angle A+\frac{1}{2}\angle B=90{}^\circ \\
& \angle 1+\angle 2=90{}^\circ ...(i) \\
\end{align}$
In $\Delta APB$
$\angle 1+\angle APB+\angle 2=180{}^\circ $
$\begin{align}
& 90{}^\circ +\angle APB=180{}^\circ [from(i)] \\
& \angle APB=90{}^\circ \\
\end{align}$
Theorem 6 : If a diagonal of a parallelogram bisects one of the angles of the parallelogram, it also bisects the second angle. Also, prove that it is a rhombus.
To Prove: AC bisects $\angle C$ and ABCD is a rhombus.
Proof: Since ABCD is a parallelogram,
AB || DC and AC intersects them
∴ $\angle 1=\angle 3...(i)$ [Alternate interior angles]
AD || BC and AC intersects them
∴ $\angle 2=\angle 4...(ii)$ [Alternate interior angles]
But AC is bisector of $\angle C$ ,
∴ $\angle 1=\angle 2...(iii)$
From (i), (ii), (iii) we get
$\angle 3=\angle 4...(iv)$
Hence, AC bisects $\angle C$
From (ii) and (iii), we have
$\angle 1=\angle 4$
BC = AB [Angles opposite to equal sides are equal]
But, AB = DC and BC= AD [ABCD is a parallelogram]
∴ AB = BC = CD = DA.
Hence, ABCD is a rhombus.
Theorem 7: The angle bisectors of a parallelogram form a rectangle.
To Prove: PQRS is a rectangle.
Proof: Since ABCD is a parallelogram,
AD || BC and transversal AB intersect them at A and B respectively
∴ $\angle A+\angle B=180{}^\circ $ [Sum of consecutive interior angles is 180].
$\begin{align}
& \frac{1}{2}\angle A+\frac{1}{2}\angle B=90{}^\circ \\
& \angle BAS+\angle ABS=90{}^\circ ...(i) \\
\end{align}$
[AS and BS are bisectors of ㄥA and ㄥB]
In $\Delta ABS$ ,
$\angle BAS+\angle ABS+\angle ASB=180{}^\circ $ [sum of the angles of a triangle is 180]
$\begin{align}
& 90{}^\circ +\angle ASB=180{}^\circ \\
& \angle ASB=90{}^\circ \\
& \angle RSP=90{}^\circ [ㄥASB and ㄥRSP are vertically opposite angles, ㄥRSP = ㄥASB ] \\
\end{align}$
Similarly, $\angle SRQ=90{}^\circ ,\angle RQP=90{}^\circ and\angle SPQ=90{}^\circ $
Hence, PQRS is a rectangle.
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