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Quadrilaterals - Class 9 : Notes


 

(1) Prove that sum of the angles of a quadrilateral is {360^ \circ }.
Given: Quadrilateral ABCD
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 …….(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 DProof: 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 = ODProof: 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 CProof: 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 BCNow, 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 COBSo, 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 = DBSo, 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   …..(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 sAOD 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 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    …(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 andDE = 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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