MATHEMATICS CLASS NOTES FOR CBSE Chapter 08. Quadrilaterals 01. Quadrilateral The word ′quad′ means four and the word ′lateral′ means sides. Thus, a plane figure bounded by four line segments AB, BC, CD and DA is called a quadrilateral and is written as quad. ABCD or, □ABCD. The points A, B, C, D are called its vertices. The four line segments, AB, BC, CD, and DA are the four sides. and the four angles ∠A, ∠B, ∠C and ∠D are the four angles of quad. ABCD. Two line segments AC and BD are called the diagonals of quad. ABCD.
Figure
02. Angle Sum Property of a Quadrilateral Result The sum of the four angles of a quadrilateral is 360°. Given : Quadrilateral ABCD To Prove : ∠A + ∠B + ∠C + ∠D = 360° Construction : Join AC
Figure
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CLASS NOTES FOR CBSE – 08. Quadrilaterals
Proof : In ∆ABC, we have ∠1 + ∠4 + ∠6 = 180° ...(i) In ∆ACD, we have ∠2 + ∠3 + ∠5 = 180° ...(i) Adding (i) and (ii), we get (∠1 + ∠2) + (∠3 + ∠4) + ∠5 + ∠6 = 180° + 180° ⇒ ∠A + ∠C + ∠D + ∠B = 360° ⇒ ∠A + ∠B + ∠C + ∠D = 360°
03. Properties of a Parallelogram Result A diagonal of parallelogram divides it into two congurent triangles. Given : A parallelogram ABCD. To Prove : A diagonal, say, AC, of parallelogram ABCD divides it into congruent triangles ABC and CDA i.e. ∆ABC ≅ ∆CDA Construction : Join AC.
Figure Proof : Since ABCD is a parallelogram. Therefore, AB || DC and AD || BC Now, AD || BC and transversal AC intersects them at A and C respectively. ∴ ∠DAC = ∠BCA [Alternate interior angles] ...(i) Again, AB||DC and trasversal AC intersects them at A and C respectively. Therefore, ∠BAC = ∠DCA [Alternate interior angles] ...(ii) Now, in ∆s ABC and CDA, we have ∠BCA = ∠DAC [From (i)] AC = AC [Common side] ∠BAC = ∠DCA [From (i)] So, by ASA congruence criterion, we obtain ∆ABC ≅ ∆CDA
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CLASS NOTES FOR CBSE – 08. Quadrilaterals
Result In a parallelogram, opposite sides are equal. Given : A parallelogram ABCD. To Prove : AB = CD and DA = BC Construction : Join AC
Figure Proof : Since ABCD is a parallelogram. Therefore, AB || DC and AD || BC Now, AD || BC and transversal AC intersects them at A and C respectively. ∴ ∠DAC = ∠BCA [Alternate interior angles] ...(i) Again, AB||DC and transversal AC intersects them at A and C respectively. ∴ ∠BAC = ∠DCA [Alternate interior angles] ...(ii) Now, in ∆s ADC and CBA, we have ∠DAC = ∠BCA [From (i)] AC = AC [Common side] and, ∠DCA = ∠BAC [From (ii)] So, by ASA criterion of congruence, we obtain ∆ADC ≅ ∆CBA ⇒ AD = CB and DC = BA [∵ Corresponding parts of congruent triangles are equal] Result The opposite angles of a parallelogram are equal. Given : A parallelogram ABCD. To Prove : ∠A = ∠C and ∠B = ∠D
Figure Proof : Since ABCD is a parallelogram. Therefore, AB || DC and AD || BC Now, AB || DC and transversal AD intersects them at A and D respectively. ∴ ∠A + ∠D = 180° [∵ Sum of consecutive interior angles is 180°] ...(i) Again, AD||BC and DC intersects them at D and C respectively. ∴ ∠D + ∠C = 180° [∵ Sum of consecutive interior angles is 180°] ...(ii)
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CLASS NOTES FOR CBSE – 08. Quadrilaterals
From (i) and (ii), we get ∠A + ∠D = ∠D + ∠C ⇒ ∠A = ∠C Similarly, ∠B = ∠D Hence, ∠A = ∠C and ∠B = ∠D Result 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
Figure Proof : Since ABCD is a parallelogram. Therefore, AB || DC and AD || BC Now, AB || DC and transversal AC intersects them at A and C respectively. ∴ ∠BAC = ∠DCA [∵Alternate interior angles are equal] ...(i) ⇒ ∠BAO = ∠DCO Again, AB||DC and BD intersects them at B and D respectively. ∴ ∠ABD = ∠CDB [∵Alternate interior angles are equal] ...(ii) ⇒ ∠ABO = ∠CDO Now, in ∆s AOB and COD, we have ∠BAO = ∠DCO [From (i)] AB = CD [∵Opposite sides of a ||gm are equal] and, ∠ABO = ∠CDO [From (ii)] So, by ASA congruence criterion ∆AOB ≅ ∆COD ⇒ OA = OC and OB = OD [c.p.c.t] Hence, OA = OC and OB = OD
04. Sufficient Conditions for a Quadrilateral to be a Parallelogram Result 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.
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CLASS NOTES FOR CBSE – 08. Quadrilaterals
Figure Proof : In ABC and CAD, we have AC = CA [Common side] CB = AD [Given] AB = CD [Given] So, by SSS criterion of congruence, we obtain ∆ACB ≅ ∆CAD ⇒ ∠CAB = ∠ACD [∵Corresponding parts of Congruent triangles are equal]...(i) and, ∠ACB = ∠CAD Now, line AC intersects AB and DC at A and C, such that ∠CAB = ∠ACD [From (i)] i.e., alternate interior angles are equal. AB || DC ...(iii) Similarly, line AC intersects BC and AD at C and A, such that ∴ ∠ACB = ∠CAD [From (i)] i.e., alternate interior angles are equal. ∴ BC || AD ...(iv) From (iii) and (iv), we have AB || DC and BC || AD. Hence, ABCD is a parallelogram. ∴ “A quadrilateral is a parallelogram if and only if its opposite sides are equal”. Result A quadrilateral is a parallelogram if its opposite angles are equal. Given : A quadrilateral ABCD in which ∠A = ∠C and ∠B = ∠D. To Prove : ABCD is a parallelogram.
Figure Proof : In quadrilateral ABCD, we have ∠A = ∠C
[Given] ...(i)
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CLASS NOTES FOR CBSE – 08. Quadrilaterals
∠B = ∠D [Given] ...(ii) ∴ ∠A + ∠B = ∠C + ∠D [Adding (i) and (ii)] ...(iii) AB = CD [Given] Since sum of the angles of a quadrilateral is 360°. ∴ ∠A + ∠B + ∠C + ∠D = 360° ...(iv) ⇒ (∠A + ∠B) + (∠A + ∠B) = 360° [Using (iii)] ⇒ 2(∠A + ∠B) = 360° ⇒ ∠A + ∠B = 180° ⇒ ∠A + ∠B = ∠C + ∠D = 180° [∵∠A + ∠B = ∠C + ∠D] ...(v) Now, line AB intersects AD and BC at A and B respectively such that ∠A + ∠B = 180° i.e., the sum of consecutive interior angles is 180° AD || BC ...(vi) Again, ∠A + ∠B = 180° ⇒ ∠C + ∠D = 180° [∵ ∠A = ∠C (Given)] Now, line BC intersects AB and DC at A and C respectively such that ∠B + ∠C = 180° i.e., the sum of consecutive interior angles is 180° ∴ AB || DC ...(vii) From (vi) and (vii), we get AD || BC and AB || DC. Hence, ABCD is a parallelogram. “A quadrilateral is a parallelogram if and only if its opposite angle are equal.” Result 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 : Quadrilateral ABCD is a parallelogram.
Figure Proof : Quadrilateral ABCD is a parallelogram. AO = OC [Given] OD = OB [Given] ∠AOD = ∠COB [Vertically opposite angles] So, by SAS criterion of congruence, we have ∆AOD ≅ ∆COB ⇒ ∠OAD = ∠OCB [∵Corresponding parts of congruent triangles are equal]...(i)
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CLASS NOTES FOR CBSE – 08. Quadrilaterals
Now, line AC intersects AD and BC at A and C respectively such that ∠OAD = ∠OCB i.e. alternate interior angles are equal. [From (i)] ∴ AD || BC Similarly, AB || CD Hence, ABCD is a parallelogram. ∴“A quadrilateral is a parallelogram if and only if its diagonals bisect each other.” Result 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 || CD. To Prove : ∆ABCD is a parallelogram. Construction : Join AC.
Figure Proof : In ∆s ABC and CDA, we have AB = DC (Given) AC = AC (Common side) and, ∠BAC = ∠DCA [∵ AB || CD Alternate interior angles are equal] So, by SAS criterion of congruence, we obtain ∆ABC ≅ ∆CDA ⇒ ∠BCA = ∠DAC [c.p.c.t] Thus, line AC intersects AB and DC at A and C respectively such that ∠DAC = ∠BCA i.e., alternate interior angles are equal. ∴ AD || CD and AD || BC. Hence, quadrilateral ABCD is a parallelogram. Summary : A quadrilateral is a parallelogram : (i) if both paris of opposite sides are equal; (ii) if both pairs of opposite angles are equal; (iii) if the diagonals bisect each other; (iv) if a pair of opposite sides are parallel as well as equal.
05. Some Useful Facts about a Triangle Result The line segment joing the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.
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CLASS NOTES FOR CBSE – 08. Quadrilaterals
Given : A ∆ABC in which D and E are the mid-points of sides AB and AC respectively. DE is joined. To Prove : DE || BC and DE = BC Construction : Produce the line segment DE to F, such that DE = EF. Join FC.
Figure Proof : In ∆s AED and CEF, we have AE = CE [∵E is the mid-point of AC] ∠AED = ∠CEF [Vertically opposite angles] and, DE = EF So, by SAS criterion of congruence, we have ∆AED ≅ ∆CEF ⇒ AD = CE [c.p.c.t.] ...(i) and, ∠ADE = ∠CFE ...(ii) Now, D is the mid-point of AB ⇒ AD = DB ⇒ DB = CF [From (i) AD = CF] ...(iii) Now, DF intersects AD and FC at D and F respectively such that ∠ADE = ∠CFE [From (ii)] i.e., alternate interior angles are equal. ∴ AD || FC ⇒ DB || 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 || BC and DF = BC [∵ Opposite sides of a ||gm are equal and parallel] But, D, E, F are collinear and DE = EF. ∴ DE || BC and DE = BC Result The line drawn through the mid-point of one side of a triangle, parallel to another side, intersects the third side at its mid-point. Given : ∆ABC in which D is the mid-point of AB and DE || BC. To Prove : E is the mid-point of AC.
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CLASS NOTES FOR CBSE – 08. Quadrilaterals
Figure Proof : We have to prove that E is the mid-point of AC. If possible, let E be not the midpoint of AC. Let E prime be the mid-point of AC. Join DE prime. Now, in ∆ABC, D is the mid-point of AB [Given] and E prime is the mid-point of AC. Therefore, above Theorem, we have DE’ || BC ...(i) Also, DE || BC [Given]...(ii) From (i) and (ii), we find that two intersecting lines DE and DE’ are both parallel to line BC. This is a contradiction to the parallel line axiom. So, our supposition is wrong. Hence, E is the mid-point of AC.
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CLASS NOTES FOR CBSE – 08. Quadrilaterals
CBSE Pattern Exercise (1) (Q 1 to 2) One Mark 1. The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral. 2. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus (Q 3 to 4) Two Marks 3. Two parallel lines l and m are intersected by a transversal p (Figure). Show that the quadrilateral formed by the bisectors of interior angles is a rectangle.
Figure 4. ABC is an isosceles triangle in which AB = AC. AD bisects exterior angle PAC and CD || AB (Figure). Show that (i) ∠DAC = ∠BCA and (ii) ABCD is a parallelogram.
Figure
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CLASS NOTES FOR CBSE – 08. Quadrilaterals
(Q 5 to 7) Three Marks 5. Diagonal AC of a parallelogram ABCD bisects ∠A (Figure). Show that
Figure (i) (ii)
it bisects ∠C also, ABCD is a rhombus.
6. ABCD is a trapezium in which AB || CD and AD = BC (Figure). Show that
Figure (i) (ii) (iii) (iv)
∠A = ∠B ∠C = ∠D ΔABC @ ΔBAD diagonal AC = diagonal BD
7. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (Figure). Show that the line segments AF and EC trisect the diagonal BD.
Figure (Q 8 to 10) Four Marks 8. ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that: (i) ABCD is a square (ii) diagonal BD bisects ∠B as well as ∠D.
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CLASS NOTES FOR CBSE – 08. Quadrilaterals
9. Show that the bisectors of angles of a parallelogram form a rectangle. 10. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.
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CLASS NOTES FOR CBSE – 08. Quadrilaterals
ANSWER
Q1 As the sum of all interior angles of a quadrilateral is 360° As the sum of all interior angles of a quadrilateral is 360°, ∴ 3x + 5x + 9x + 13x = 360° 30x = 360° x = 12° Hence, the angles are 3x = 3 × 12 = 36° 5x = 5 × 12 = 60° 9x = 9 × 12 = 108° 13x = 13 × 12 = 1560° Q2
Figure Let ABCD be a quadrilateral, whose diagonals AC and BD bisect each other at right angle i.e., OA = OC, OB = OD, and ∠AOB = ∠BOC = ∠COD = ∠AOD = 90°. To prove ABCD a rhombus, we have to prove ABCD is a parallelogram and all the sides of ABCD are equal. In ∆AOD and ∆COD, OA = OC
(Diagonals bisect each other)
∠AOD = ∠COD
(Given)
OD = OD
(Common)
∴ ∆AOD ≅ ∆COD
(By SAS congruence rule)
∴ AD = CD
...(i)
Similary, it can be proved that AD = AB and CD = BC
...(ii)
From equations (i) and (ii), AB = BC = CD = AD since opposite sides of quadrilateral ABCD are equal, it can be said that ABCD is a parallelogram. since all sides of a parallelogram ABCD are equal, it can be said that ABCD is a rhombus.
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CLASS NOTES FOR CBSE – 08. Quadrilaterals
Q3 It is given that PS || QR and transversal p intersects them at points A and C respectively. The bisectors of ∠PAC and ∠ACQ intersect at B and bisectors of ∠ACR and ∠SAC intersect at D. We are to show that quadrilateral ABCD is a rectangle. Now,
∠PAC = ∠ACR
So,
∠PAC = ∠ACR
i.e.,
∠BAC = ∠ACD
(Alternate angles as l || m and p is a transversal)
These form a pair of alternate angles for lines AB and DC with AC as transversal and they are equal also. So,
AB || DC
Similarly,
BC || AD
(Considering ∠ACB = ∠CAD)
Therefore, quadrilateral ABCD is a parallelogram. Also,
∠PAC = ∠CAS = 180° (Linear pair)
So,
∠PAC + ∠CAS = × 180° = 90°
or,
∠BAC = ∠CAD = 90°
or,
∠BAD = 90°
So, ABCD is a parallelogram in which one angle is 90°. Therefore, ABCD is a rectangle. Q4 (i)
∆ABC is isosceles in which AB = AC
(Given)
So,
∠ABC = ∠ACB
(Angles opposite to equal sides)
Also,
∠PAC = ∠ABC + ∠ACB
(Exterior angle of a triangle)
or,
∠PAC = 2∠ACB
...(i)
Now, AD bisects ∠PAC. So,
∠PAC = 2∠DAC
...(ii)
2∠DAC = 2∠ACB
[From (i) and (ii)]
Therefore, ∠DAC = ∠ACB
or, (ii)
Now, these equal angles form a pair of alternate angles when line segments BC and AD are intersected by a transversal AC. So,
BC || AD
Also,
BA || CD
(Given)
Now, both pairs of opposite sides of quadrilateral ABCD are parallel. So, ABCD is a parallelogram.
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CLASS NOTES FOR CBSE – 08. Quadrilaterals
Q5 (i)
ABCD is a parallelogram. ∠DAC = ∠BCA
(Alternate interior angles) ...(i)
And, ∠BAC = ∠DCA
(Alternate interior angles) ...(ii)
However, it is given that AC bisects ∠A. ∠DAC = ∠BAC
...(iii)
From equations (i), (ii) and (iii), we obtain ∠DAC = ∠BAC = ∠BAC = ∠DCA
,...(iv)
∠DCA = ∠BCA Hence, AC bisects ∠C. (ii)
From equation (iv) we obtain ∠DAC = ∠DCA DA = DC
(Side opposite to equal angles are equal)
However, DA = BC and AB = CD
(Opposite sides of a parallelogram)
AB = BC = CD = DA Hence, ABCD is a rhombus. Q6
Figure Let us extend AB. Then, draw a line through C, which is parallel to AD, intersecting AE at point E. It is clear that AECD is a parallelogram. (i)
AD = CE
(Opposite sides of parallelogram AECD)
However, AD = BC
(Given)
Therefore, BC = CE
∠CEB = ∠CBE
(Angle opposite to equal sides are also equal)
Consider parallel lines AD and CE. AE is the transversal line for them. ∠A + ∠CEB = 180°
(Angles on the same side of transversal)
∠A+ ∠CBE = 180°
(Using the relation ∠CEB = ∠CBE) ...(i)
However, ∠B + ∠CBE = 180°
(Linear pair angles) ...(ii)
From equations (i) and (ii), we obtain ∠A = ∠B
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CLASS NOTES FOR CBSE – 08. Quadrilaterals
(ii)
AB || CD ∠A + ∠D = 180°
(Angles on the same side of the transversal)
Also, ∠C + ∠B = 180°
(Angles on the same side of the transversal)
∠A + ∠D = ∠C + ∠B However, ∠A = ∠B
[Using the result obtained in (i)]
∠C = ∠D (iii) In ∆ABC and ∆BAD, AB = BA
(Common side)
BC = AD
(Given)
∠B = ∠A
(Proved before)
∆ABC ≅ ∆BAD
(SAS congruence rule)
(iv) We had observed that, ∆ABC ≅ ∆BAD AC = BD
(By c.p.c.t.)
Q7 ABCD is a parallelogram. AB || CD And hence, AE || FC Again, AB = CD
(Opposite sides of parallelogram ABCD)
AB = CD AE = FC
(E and F are mid-points of side AB and CD)
In quadrilateral AECF, one pair of opposite sides (AE and CF) is parallel and equal to each other. Therefore, AECF is a parallelogram. AF || EC
(Opposite sides of a parallelogram)
In ∆DQC, F is the mid-point of side DC and FP || CQ (as AF || EC). Therefore, by using the converse of mid-point theorem, it can be said that P is the mid-point of DQ. DP = PQ
...(i)
Similarly, in ∆APB, E is the mid-point of side AB and EQ || AP (as AF || EC). Therefore, by using the converse of mid-point theorem, it can be said that Q is the mid-point of PB. PQ = QB
...(ii)
From equations (i) and (ii), DP = PQ = BQ Hence, the line segments AF and EC trisect the diagonal BD.
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CLASS NOTES FOR CBSE – 08. Quadrilaterals
Q8
Figure (i)
It is given that ABCD is a rectangle. ∠A = ∠C ⇒ ∠A = ∠C ⇒ ∠DAC = ∠DCA
(AC bisects ∠A and ∠C)
CD = DA
(Sides opposite to equal angles are also equal)
However, DA = BC and AB = CD
(Opposite sides of a rectangle are equal)
AB = BC = CD = DA ABCD is a rectangle and all of its sides are equal. Hence, ABCD is a square. (ii)
Let us join BD. In ∆BCD, BC = CD
(Sides of a square are equal to each other)
∠CDB = ∠CBD
(Angles opposite to equal sides are equal)
However, ∠CDB = ∠ABD
(Alternate interior angles for AB || CD)
∠CBD = ∠ABD ∠BD bisects ∠B. Also, ∠CBD = ∠ADB
(Alternate interior angles for BC || AD)
∠CDB = ∠ABD ∠BD bisects ∠D Q9
Figure Let P, Q, R and S be the points of intersection of the bisectors of ∠A and ∠B, ∠B, ∠C, ∠C and ∠D, and ∠D and ∠A respectively of parallelogram ABCD (above figure).
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CLASS NOTES FOR CBSE – 08. Quadrilaterals
In ∆ASD, what do you observe ? Since DS bisects ∠D and AS bisects ∠A, therefore, ∠DAS + ∠ADS = ∠A + ∠D = (∠A + ∠D) (∠A and ∠D are interior angles on the same side of transversal) = × 180°
(∠A and ∠D are interior angles on the same side of transversal)
= 90° Also, ∠DAS + ∠ADS + ∠DAS = 180° or,
90° + ∠DSA = 180°
or,
∠DSA = 90°
So,
∠PSR = 90°
(Angle sum property of a triangle)
(Being vertically opposite to ∠DSA)
Similarly, it can be shown that ∠APB = 90° or ∠SPQ = 90° (as it was shown for ∠DSA). Similarly, ∠PQR = 90° and ∠SRQ = 90°. So, PQRS is a quadrilateral in which all angles are right angles.∠PSR = ∠PQR =90° and ∠SPQ = ∠SRQ = 90°. So both pairs of opposite angles are equal. Therefore, PQRS is a parallelogram in which one angle (in fact all angles) is 90° and so, PQRS is a rectangle. Q10
Figure In ∆ABC, P and Q are the mid-points of sides AB and BC respectively. PQ || AC and PQ = AC
(Using mid-point theorem) ...(i)
In ∆ABC, R and S are the mid-points of CD and AD respectively. RS || AC and RS = AC
(Using mid-point theorem) ...(ii)
From equations ...(i) and ...(ii), we obtain PQ || RS and PQ = RS Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other, it is a parallelogram.
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CLASS NOTES FOR CBSE – 08. Quadrilaterals
Let the diagonals of rhombus ABCD intersect each other at point O. In quadrilateral OMQN, MQ || ON
(∵ PQ || AC)
QN || OM
(∵ QR || BD)
Therefore, OMQN is a parallelogram. ∠MQN = ∠NOM ∠PQR = ∠NOM However, ∠NOM = 90°
(Diagonals of a rhombus are perpendicular to each other)
∠PQR = 90° Clearly, PQRS is a parallelogram having one of its interior angles as 90°. Hence, PQRS is a rectangle.
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