# Quadrilaterals - Class 9 : Notes

(1) Prove that sum of the angles of a quadrilateral is ${360^ \circ }$.
To Prove: $\angle A + \angle B + \angle C + \angle D = {360^ \circ }$
Construction: Join ACProof: In $\Delta ABC$, We have
$\angle 1 + \angle 4 + \angle 6 = {180^ \circ }$..........(i)
In $\Delta ACD$, we have
$\angle 2 + \angle 3 + \angle 5 = {180^ \circ }$...............(ii)
Adding (i) and (ii), we get
$\left( {\angle 1 + \angle 2} \right) + \left( {\angle 3 + \angle 4} \right) + \left( {\angle 5 + \angle 6} \right) = {180^ \circ } + {180^ \circ }$
$\angle A + \angle C + \angle D + \angle B = {360^ \circ }$
$\angle A + \angle B + \angle C + \angle D = {360^ \circ }$

(2) Prove that a diagonal of a parallelogram divides it into two congruent triangles.
Given: A parallelogram ABCD
To Prove: A diagonal, say, AC, of parallelogram ABCD divides it into congruent triangles ABC and CDA i.e.
$\Delta ABC \cong \Delta CDA$
Construction: Join ACProof: Since ABCD is a parallelogram. Therefore,
$AB\parallel DC$ and $AD\parallel BC$
Now, $AD\parallel BC$ and transversal AC intersects them at A and C respectively.
$\angle DAC = \angle BCA$ â€¦â€¦.(i)Â Â Â Â Â Â Â Â Â Â Â Â Â Â  Â  [Alternate interior angles]
Again, $AB\parallel DC$ and transversal AC intersects them at A and C respectively. Therefore,
$\angle BAC = \angle DCA$Â  â€¦â€¦(ii) Â  Â  Â  Â  Â  Â  Â  Â [Alternate interior angles]
Now, in $\Delta s$ ABC and CDA, we have
$\angle BCA = \angle DAC$Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  [From (i)]
$AC = AC$
$\angle BAC = \angle DCA$
So, by ASA congruence criterion, we have
$\Delta ABC \cong \Delta CDA$

(3) Prove that two 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. Therefore,
$AB\parallel DC$ and $AD\parallel BC$
Now, $AB\parallel DC$ and transversal AD intersects them at A and D respectively.
$\angle A + \angle D = {180^ \circ }$Â  â€¦â€¦.(i)Â Â  [Sum of Consecutive interior anglesis ${180^ \circ }$ ]
Again, $AD\parallel BC$ and DC intersects them at D and C respectively.
$\angle D + \angle C = {180^ \circ }$ â€¦.. (ii)Â Â Â Â  [Sum of Consecutive interior angles is ${180^ \circ }$ ]
From (i) and (ii), we get
$\angle A + \angle D = \angle D + \angle C$
$\angle A = \angle C$.
Similarly, $\angle B = \angle D$.
Hence, $\angle A = \angle C$ and $\angle B = \angle D$

(4) Prove that 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. Therefore,
$AB\parallel DC$ and $AD\parallel BC$
Now, $AB\parallel DC$ and transversal AC intersects them at A and C respectively.
$\angle BAC = \angle DCA$
$\angle BAO = \angle DCO$ â€¦â€¦..(i)
Again, $AB\parallel DC$ and BD intersects them at B and D respectively.
$\angle ABD = \angle CDB$
$\angle ABO = \angle CDO$ â€¦â€¦..(ii)
Now, in $\Delta s$ AOB and COD, we have
$\angle BAO = \angle DCO$
$AB = CD$
and, $\angle ABO = \angle CDO$
So, by ASA congruence criterion
$\Delta AOB \cong \Delta COD$
$OA = OC$ and $OB = OD$
Hence, $OA = OC$ and $OB = OD$

(5) Prove that 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. Therefore,
$AD\parallel BC$
Now, $AD\parallel BC$ and transversal AB intersects them.
$\angle A + \angle B = {180^\circ }$
$\frac{1}{2}\angle A + \frac{1}{2}\angle B = {90^\circ }$
$\angle 1 + \angle 2 = {90^ \circ }$ â€¦.(i)
AP is the bisector of $\angle A$ and BP is the bisector of $\angle B$ then $\angle 1 = \frac{1}{2}\angle A$Â and $\angle 2 = \frac{1}{2}\angle B$
In $\Delta APB$, we have
$\angle 1 + \angle APB + \angle 2 = {180^ \circ }$
${90^ \circ } + \angle APB = {180^ \circ }$ Â  Â  Â  Â  Â [From (i)]
$\angle APB = {90^ \circ }$

(6) Prove that if a diagonal of a parallelogram bisects one of the angles of the parallelogram it also bisects the second angle.
Given: A parallelogram ABCD in which diagonal AC bisects $\angle A$.
To prove: AC bisects $\angle C$Proof: Since ABCD is a parallelogram. Therefore,
$AB\parallel DC$
Now, $AB\parallel DC$ and AC intersects them.
$\angle 1 = \angle 3$Â Â  â€¦â€¦(i) [Alternate interior angles]
Again, $AD\parallel BC$ and AC intersects them.
$\angle 2 = \angle 4$Â  â€¦â€¦(ii) [Alternate interior angles]
But, it is given that AC is the bisector of $\angle A$. Therefore,
$\angle 1 = \angle 2$Â Â Â Â Â Â Â Â  â€¦â€¦..(iii)
From (i), (ii) and (iii), we get
$\angle 3 = \angle 4$Â Â Â Â Â Â Â  â€¦â€¦â€¦(iv)
Hence, AC bisects $\angle C$.
From (ii) and (iii), we have
$\angle 1 = \angle 4$
$BC = AB$Â Â Â Â Â  [Angles opposite to equal sides are equal]
But, $AB = DC$ and $BC = AD$Â Â Â  [ ABCD is a parallelogram]
$AB = BC = CD = DA$
Hence, ABCD is a rhombus.

(7) Prove that the angles bisectors of a parallelogram form a rectangle.
Proof: Since ABCD is a parallelogram. Therefore,
$AD\parallel BC$Now, $AD\parallel BC$ and transversal AB intersects them at A and B respectively. Therefore,
$\angle A + \angle B = {180^ \circ }$Â  [Sum of consecutive interior angles is ${180^ \circ }$]
$\frac{1}{2}\angle A + \frac{1}{2}\angle B = {90^ \circ }$
$\angle BAS + \angle ABS = {90^ \circ }$Â  â€¦.(i)Â Â  [AS and BS are bisectors of $\angle A$ and $\angle B$ respectively]
But, in $\Delta ABS$, we have
$\angle BAS + \angle ABS + \angle ASB = {180^ \circ }$ [Sum of the angle of a triangle is ${180^ \circ }$]
${90^\circ } + \angle ASB = {180^\circ }$
$\angle ASB = {90^ \circ }$
$\angle RSP = {90^ \circ }$ [$\angle ASB$ and $\angle RSP$ are vertically opposite angles $\angle RSP = \angle ASB$]
Similarly, we can prove that
$\angle SRQ = {90^ \circ }$, $\angle RQP = {90^ \circ }$ and $\angle SPQ = {90^ \circ }$
Hence, PQRS is a rectangle.

(8) Prove that 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 $\Delta s$ ACB and CAD, we have
$AC = CA$Â Â  [Common Side]
$CB = AD$
$AB = CD$
So, by SAS criterion of congruence, we have
$\Delta s$ ACB and CAD
$\angle CAB = \angle ACD$ Â  Â  Â  Â â€¦.(i)
And, $\angle ACB = \angle CAD$
Now, line AC intersects AB and DC at A and C, such that
$\angle CAB = \angle ACD$ Â  Â  Â .....(ii)
i.e., alternate interior angles are equal.
$AB\parallel DC$Â Â Â  â€¦..(iii)
Similarly, line AC intersects BC and AD at C and A such that
$\angle ACB = \angle CAD$
i.e., alternate interior angles are equal.
$BC\parallel AD$Â Â Â  â€¦..(iv)
From (iii) and (iv), we have
$AB\parallel DC$ and $BC\parallel AD$
Hence, ABCD is a parallelogram.

(9) Prove that a quadrilateral is a parallelogram if its opposite angles are equal.
Given: A quadrilateral ABCD in which $\angle A = \angle C$ and $\angle B = \angle D$.
To prove: ABCD is a parallelogram.Proof: In quadrilateral ABCD, we have
$\angle A = \angle C$ â€¦â€¦(i)
$\angle B = \angle D$Â Â  â€¦â€¦(ii)
$\angle A + \angle B = \angle C + \angle D$Â  â€¦..(iii)
Since sum of the angles of a quadrilateral is ${360^ \circ }$
$\angle A + \angle B + \angle C + \angle D = {360^ \circ }$ Â  â€¦â€¦(iv)
$\left( {\angle A + \angle B} \right) + \left( {\angle A + \angle B} \right) = {\text{ }}360^\circ$
$2\left( {\angle A + \angle B} \right) = {360^\circ }$
$\left( {\angle A + \angle B} \right) = {180^ \circ }$
$\angle A + \angle B = \angle C + \angle D = {180^ \circ }$Â  â€¦..(v) Â  Â [${\angle A + \angle B = \angle C + \angle D}$]
Now, line AB intersects AD and BC at A and B respectively such that
$\angle A + \angle B = {180^ \circ }$
i.e. the sum of consecutive interior angles is ${180^ \circ }$
$AD\parallel BC$Â Â Â  â€¦â€¦(vi)
Again, $\angle A + \angle B = {180^ \circ }$
$\angle C + \angle B = {180^ \circ }$
Now, line BC intersects AB and DC at A and C respectively such that
$\angle B + \angle C = {180^ \circ }$
i.e., the sum of consecutive interior angles is ${180^ \circ }$.
$AB\parallel DC$Â Â Â Â  â€¦â€¦(vii)
From (vi) and (vii), we get
$AD\parallel BC$ and $AB\parallel DC$.
Hence, ABCD is a parallelogram.

(10) Prove that the diagonals of a quadrilateral bisect each other, if it is a parallelogram.
Proof: In $\Delta s$ AOD and COB, we have
$AO = OC$
$OD = OB$
$\angle AOD = \angle COB$So, by SAS criterion of congruence, we have
$\Delta AOD \cong \Delta COB$
$\angle OAD = \angle OCB$
Now, line AC intersects AD and BC at A and C respectively such that
$\angle OAD = \angle OCB$
i.e., alternate interior angles are equal.
$AD\parallel BC$
Similarly, $AB\parallel CD$
Hence, ABCD is a parallelogram.

(11) Prove that a quadrilateral is a parallelogram if its one Pair of opposite sides are equal and parallel.
Given: A quadrilateral ABCD in which $AB = CD$ and $AB\parallel CD$.
To prove: $\Delta ABCD$ is a parallelogram.
Construction: Join AC.Proof: In $\Delta s$ ABC and CDA, we have
$AB = DC$
$AC = AC$
And, $\angle BAC = \angle DCA$
So, by SAS criterion of congruence, we have
$\Delta ABC \cong \Delta CDA$
$\angle BCA = \angle DAC$
Thus, line AC intersects AB and DC at A and C respectively such that
$\angle DAC = \angle BCA$
i.e., alternate interior angles are equal.
$AD\parallel BC$ .
Thus, $AB\parallel CD$ and $AD\parallel BC$.
Hence, quadrilateral ABCD is a parallelogram.

(12) Prove that each of the four angles of a rectangle is a right angle.
Given: A rectangle ABCD such that $\angle A = {90^ \circ }$
To prove: $\angle A = \angle B = \angle C = \angle D = {90^ \circ }$Proof: Since ABCD is a rectangle.
ABCD is a parallelogram
$AD\parallel BC$
Now, $AD\parallel BC$ and line AB intersects them at A and B.
$\angle A + \angle B = {180^ \circ }$
${90^ \circ } + \angle B = {180^ \circ }$
$\angle B = {90^ \circ }$
Similarly, we can show that $\angle C = {90^ \circ }$ and $\angle D = {90^ \circ }$
Hence, $\angle A = \angle B = \angle C = \angle D = {90^ \circ }$

(13) Prove that each of the four sides of a rhombus of the same length.
Given: A rhombus ABCD such that $AB = BC$.
To prove: $AB = BC = CD = DA$.Proof: Since ABCD is rhombus
ABCD is a parallelogram
$AB = CD$ and $BC = AD$
But, $AB = BC$
$AB = BC = CD = DA$
Hence, all the four sides of a rhombus are equal.

(14) Prove that the diagonals of a rectangle are of equal length.
Given: A rectangle ABCD with AC and BD as its diagonals.
To prove: $AC = BD$.Proof: Since ABCD is a rectangle
ABCD is a parallelogram such that one of its angles, say , $\angle A$ is a right angle.
$AD = BC$ and $\angle A = {90^ \circ }$
Now, $AD\parallel BC$ and AB intersects them at A and B respectively.
$\angle A + \angle B = {180^ \circ }$
${90^ \circ } + \angle B = {180^ \circ }$
$\angle B = {90^ \circ }$]
In $\Delta s$ ABD and BAC, we have
$AB = BA$
$\angle A = \angle B$
And, $AD = BC$
So, by SAS criterion of congruence, we have
$\Delta ABD \cong \Delta BAC$}
$BD = AC$
Hence, $AC = BD$

(15) Prove that diagonals of a parallelogram are equal if and only if it is a rectangle.
Proof: In $\Delta s$ ABC and DCB, we have
$AB = DC$
$BC = CB$
And, $AC = DB$So, by SAS criterion of congruence, we have
$\Delta ABC \cong \Delta DCB$
$\angle ABC = \angle DCB$
But, $AB\parallel DC$ and BC cuts them.
$\angle ABC + \angle DCB = {180^ \circ }$
$2\angle ABC = {180^ \circ }$
$\angle ABC = {90^ \circ }$
Thus, $\angle ABC = \angle DCB = {90^ \circ }$.
Hence, ABCD is a rectangle.

(16) Prove that the diagonals of a rhombus are perpendicular to each other.
Given: A rhombus ABCD whose diagonals AC and BD intersect at O.
To prove: $\angle BOC = \angle DOC = \angle AOD = \angle AOB = {90^ \circ }$4Proof: We know that a parallelogram is a rhombus, if ll of its sides are equal. So, ABCD is a rhombus. This implies that ABCD is a parallelogram such that
$AB = BC = CD = DA$Â Â  â€¦..(i)
Since the diagonals of a parallelogram bisect each other.
$OB = OD$ and $OA = OC$Â Â Â  â€¦..(ii)
Now, in $\Delta s$ BOC and DOC, we have
$BO = OD$
$BC = DC$
$OC = OC$
So, by SSS criterion of congruence, we have
$\Delta BOC \cong \Delta DOC$
$\angle BOC = \angle DOC$
But, $\angle BOC + \angle DOC = {180^ \circ }$
$\angle BOC = \angle DOC = {90^ \circ }$
Similarly, $\angle AOB = \angle AOD = {90^ \circ }$
Hence, $\angle AOB = \angle BOC = \angle COD = \angle DOA = {90^ \circ }$

(17) Prove that diagonals of a parallelogram are perpendicular if and only if it is a rhombus.
Given: A parallelogram ABCD in which $AC \bot BD$.
To prove: Parallelogram ABCD is a rhombus.Proof: Suppose AC and BD intersect at O. Since the diagonals of a parallelogram bisect each other. So, we have
$OA = OC$Â Â  â€¦â€¦(i)
Now, in $\Delta s$AOD and COD, we have
$OA = OC$
$\angle AOD \cong \angle COD$
$OD = OD$
So, by SAS criterion of congruence, we have
$\Delta AOD \cong \Delta COD$
$AD = CD$Â Â Â  â€¦â€¦(ii)
Since ABCD is a parallelogram.
$AB = CD$ and $AD = CD$
$AB = CD = AD = BC$
Hence, parallelogram ABCD is a rhombus.

(18) Prove that the diagonals of a square are equal and perpendicular to each other.
Given: A square ABCD.
To prove: $AC = BD$ and $AC \bot BD$.Proof: In $\Delta s$ ADB and BCA, we have
$AD = BC$
$\angle BAD = \angle ABC$
And, $AB = BA$
So, by SAS criterion of congruence, we have
$\Delta ADB \cong \Delta BCA$
$AC = BD$
Now, in $\Delta s$ AOB and AOD, we have
$OB = OD$
$AB = AD$
And, $AO = AO$
So, by SSS criterion of congruence, we have
$\Delta AOB \cong \Delta AOD$
$\angle AOB = \angle AOD$
But, $\angle AOB + \angle AOD = {180^ \circ }$
$\angle AOB = \angle AOD = {90^ \circ }$
$AO \bot BD$
$AC \bot BD$
Hence, $AC = BD$ and $AC \bot BD$

(19) Prove that if the diagonals of a parallelogram are equal and intersect at right angles, Â then it is square
Given: A parallelogram ABCD in which $AC = BD$ and $AC \bot BD$
To prove: ABCD is a square.Proof: In $\Delta s$ AOB and AOD, we have
$AO = AO$
$\angle AOB = \angle AOD$
And $OB = OD$
So, by SAS criterion of congruence, we have
$\Delta AOB \cong \Delta AOD$
$AB = AD$
But, $AB = CD$ and $AD = BC$
$AB = BC = CD = DA$Â Â  â€¦â€¦.(i)
Now, in $\Delta s$ ABD and BAC, we have
$AB = BA$
$AD = BC$
And, $BD = AC$
So, by SSS criterion of congruence, we have
$\Delta ABD \cong \Delta BAC$
$\angle DAB = \angle CBA$
But, $\angle DAB + \angle CBA = {180^\circ }$
$\angle DAB = \angle CBA = {90^ \circ }$Â Â  â€¦â€¦(ii)
From (i) and (ii), we obtain that ABCD is a parallelogram whose all side are equal and all angles are right angles.
Hence, ABCD is a square.

(20) Prove that the line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.
Given: A $\Delta ABC$ in which D and E are the mid points of sides AB and AC respectively. DE is joined
To prove: $DE\parallel BC$ and $DE = \frac{1}{2}BC$
Construction: Produce the line segment DE to F, such that $DE = EF$. Join FCProof: In $\Delta s$ AED and CEF, we have
$AE = CE$
$\angle AED = \angle CEF$
And, $DE = EF$
So, by SAS criterion of congruence, we have
$\Delta AED \cong \Delta CFE$
$AD\parallel CF$Â Â Â  â€¦(i)
And, $\angle ADE = \angle CFE$Â  â€¦â€¦(ii)
Now, D is the mid-point of AB
$AD = DB$
$DB = CF$ Â Â â€¦..(iii)
Now, DF intersects AD and FC at D and F respectively such that
$\angle ADE = \angle CFE$
i.e. alternate interior angles are equal.
$AD\parallel FC$
$DB\parallel CF$Â Â Â  â€¦..(iv)
From (iii) and (iv), we find that DBCF is a quadrilateral such that one pair of sides are equal and parallel.
DBCF is a parallelogram.
$DF\parallel BC$ and $DF = BC$
But, D,E,F are collinear and$DE = EF$.
$DE\parallel BC$ and $DE = \frac{1}{2}BC$

(21) Prove that a line through the mid-point of a side of a triangle parallel to another side bisects theÂ third side.
Proof: We have to prove that E is the mid-point of AC. If possible, let E be not the mid-point of AC. Let E prime be the mid-point AC. Join DE prime.Now, in $\Delta ABC$, D is the mid-point of AB and E prime is the mid-point of AC. We have,
$DE'\parallel BC$Â Â  â€¦..(i)
Also, $DE\parallel BC$Â  â€¦.(ii)
From (i) and (ii), we find that two intersecting lines DE and DEâ€™ are both parallel to Line BC.
This is contradiction to the parallel line axiom.
So, our supposition is wrong. Hence, E is the mid-point of AC.

(22) Prove that the quadrilateral formed by joining the mid-points of the sides of a quadrilateral, in order, is a parallelogram.
Given: ABCD is a quadrilateral in which P,Q,R,S are the mid-points of sides AB, BC, CD and DA respectively.
To prove: PQRS is a parallelogram.Proof: In $\Delta ABC$, P and Q are the mid-points of sides AB and BC respectively.
$PQ\parallel AC$ and $PQ = \frac{1}{2}AC$Â Â  â€¦.(i)
In $\Delta ADC$, R and S are the mid-points of CD and AD respectively.
$RS\parallel AC$ and $RS = \frac{1}{2}AC$Â Â Â  â€¦â€¦.(ii)
From (i) and (ii), we have
$PQ = RS$ and $PQ\parallel RS$
Thus, in quadrilateral PQRS one pair of opposite sides are equal and parallel.
Hence, PQRS is a parallelogram.

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