# 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 AC Proof: 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 AC Proof: 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.
$\angle DAC = \angle BCA$ …….(i)                 [Alternate interior angles]
Again, $AB\parallel DC$ and transversal AC intersects them at A and C respectively. Therefore,
$\angle BAC = \angle DCA$  ……(ii)                [Alternate interior angles]
Now, in $\Delta s$ ABC and CDA, we have
$\angle BCA = \angle DAC$                                            [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.
$\angle A + \angle D = {180^ \circ }$  …….(i)   [Sum of Consecutive interior anglesis ${180^ \circ }$ ]
Again, $AD\parallel BC$ and DC intersects them at D and C respectively.
$\angle D + \angle C = {180^ \circ }$ ….. (ii)     [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.
$\angle BAC = \angle DCA$
$\angle BAO = \angle DCO$ ……..(i)
Again, $AB\parallel DC$ and BD intersects them at B and D respectively.
$\angle ABD = \angle CDB$
$\angle ABO = \angle CDO$ ……..(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.
$\angle A + \angle B = {180^\circ }$
$\frac{1}{2}\angle A + \frac{1}{2}\angle B = {90^\circ }$
$\angle 1 + \angle 2 = {90^ \circ }$ ….(i)
AP is the bisector of $\angle A$ and BP is the bisector of $\angle B$ then $\angle 1 = \frac{1}{2}\angle A$ and $\angle 2 = \frac{1}{2}\angle B$
In $\Delta APB$, we have
$\angle 1 + \angle APB + \angle 2 = {180^ \circ }$
${90^ \circ } + \angle APB = {180^ \circ }$          [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.
$\angle 1 = \angle 3$   ……(i) [Alternate interior angles]
Again, $AD\parallel BC$ and AC intersects them.
$\angle 2 = \angle 4$  ……(ii) [Alternate interior angles]
But, it is given that AC is the bisector of $\angle A$. Therefore,
$\angle 1 = \angle 2$         ……..(iii)
From (i), (ii) and (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.

(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,
$\angle A + \angle B = {180^ \circ }$  [Sum of consecutive interior angles is ${180^ \circ }$]
$\frac{1}{2}\angle A + \frac{1}{2}\angle B = {90^ \circ }$
$\angle BAS + \angle ABS = {90^ \circ }$  ….(i)   [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
$AC = CA$   [Common Side]
$CB = AD$
$AB = CD$
So, by SAS criterion of congruence, we have
$\Delta s$ ACB and CAD
$\angle CAB = \angle ACD$        ….(i)
And, $\angle ACB = \angle CAD$
Now, line AC intersects AB and DC at A and C, such that
$\angle CAB = \angle ACD$      .....(ii)
i.e., alternate interior angles are equal.
$AB\parallel DC$    …..(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.
$BC\parallel AD$    …..(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
$\angle A = \angle C$ ……(i)
$\angle B = \angle D$   ……(ii)
$\angle A + \angle B = \angle C + \angle D$  …..(iii)
Since sum of the angles of a quadrilateral is ${360^ \circ }$
$\angle A + \angle B + \angle C + \angle D = {360^ \circ }$   ……(iv)
$\left( {\angle A + \angle B} \right) + \left( {\angle A + \angle B} \right) = {\text{ }}360^\circ$
$2\left( {\angle A + \angle B} \right) = {360^\circ }$
$\left( {\angle A + \angle B} \right) = {180^ \circ }$
$\angle A + \angle B = \angle C + \angle D = {180^ \circ }$  …..(v)    [${\angle A + \angle B = \angle C + \angle D}$]
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 ${180^ \circ }$
$AD\parallel BC$    ……(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 ${180^ \circ }$.
$AB\parallel DC$     ……(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 }$4 Proof: 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
$AB = BC = CD = DA$   …..(i)
Since the diagonals of a parallelogram bisect each other.
$OB = OD$ and $OA = OC$    …..(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
$OA = OC$   ……(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
$\Delta AOD \cong \Delta COD$
$AD = CD$    ……(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,  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 $AD = BC$
$AB = BC = CD = DA$   …….(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, $\angle DAB + \angle CBA = {180^\circ }$
$\angle DAB = \angle CBA = {90^ \circ }$   ……(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 FC Proof: 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
$\Delta AED \cong \Delta CFE$
$AD\parallel CF$    …(i)
And, $\angle ADE = \angle CFE$  ……(ii)
Now, D is the mid-point of AB
$AD = DB$
$DB = CF$   …..(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.
$AD\parallel FC$
$DB\parallel CF$    …..(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 the 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,
$DE'\parallel BC$   …..(i)
Also, $DE\parallel BC$  ….(ii)
From (i) and (ii), we find that two intersecting lines DE and DE’ 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 $PQ = \frac{1}{2}AC$   ….(i)
In $\Delta ADC$, R and S are the mid-points of CD and AD respectively.
$RS\parallel AC$ and $RS = \frac{1}{2}AC$    …….(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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