NCERT Solutions for Class 9 Maths Chapter 8: Quadrilaterals

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    Q. ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.

    Ans. Given : ABCD is a rhombus. 
    AB = BC = CD = AD
    Now, in △ ABC and △ ADC,


    BC = AD [Given]
    AC = AC [Common side]
    AB = DC [Given]
    △ ABC △ ADC [By SSS]
    ∠BAC = ∠DAC
    and ∠ACB = ∠ACD [CPCT]
    Hence, AC bisects ∠A as well as ∠C.
    Similarly , △ BAD △ BCD
    ∠ABD = ∠DBC and ∠ADB = ∠BDC
    Hence, BD bisects ∠B as well as ∠D.

    Q. D, E and F are respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. Prove that DDEF is also an equilateral triangle.

    Ans. Given : ABC is an equilateral triangle in which D, E, F are the mid-points of side, BC, AC, AB, respectively.
    To prove : DEF is an equilateral triangle.
    Proof : We have, F and E are the mid-points of sides AB and AC, respectively.

    equilateral triangle 1

    $$\text{By mid-point theorem,} \\FE=\frac{1}{2}BC…(i) \\\text{Similarly , E and D are the mid-points of AC and BC, respectively.} \\ ED=\frac{1}{2}AB…(ii) \\ \text{And, F and D are the mid-points of AB and BC respectively.} \\ FD=\frac{1}{2}AC…(iii) \\ \text{Now, ABC is an equilateral triangle} \\ AB = BC = CA \\or\space \frac{1}{2}AB=\frac{1}{2}BC=\frac{1}{2}CA \\ ⇒ ED = FE = FD [Using (i), (ii) and (iii)] \\ \bigtriangleup DEF\space text{is an equilateral triangle. Hence Proved.} $$

    Q. In figure, ABCD is a trapezium in which side AB is parallel to side DC and E is the mid-point of side AD. F is a point on the side BC such that the segment EF is parallel to side DC.

    Ans. Given : ABCD is a trapezium with AB || DC. Also E is the mid-point of AD such that EF || DC.
    AB || EF || DC.
    trapezium 1

    $$\text{To prove :}EX=\frac{1}{2}(AB+DC)\\ \text{Construction : Join BD such that it intersect EF at X.}\\ \text{Proof : In DABD, E is the mid-point of AD and EX || AB.}\\ \text{By the converse of mid-point theorem,}\\ \text{X is the mid-point of DB.}\\ \text{and}\space EX=\frac{1}{2}AB….(i)\\ \text{Similarly, in DBDC, X is the mid-point of BD and XF || DC.}\\ \text{F is the mid-point of BC.}\\ \text{and}\space XX=\frac{1}{2}DC….(ii)\\ \text{Adding (i) and (ii), we get}\\ EX + XF =\frac{1}{2}AB+\frac{1}{2}AD\\ ⇒ EF =\frac{1}{2}(AB+DC)$$

    Q. Prove that the angles bisectors of a parallelogram form a rectangle.

    Ans. Consider a parallelogram ABCD in which AP, BP, CR and
    DR are bisectors at ∠A, ∠B, ∠C and ∠D respectively.
    To prove : PQRS is a rectangle.
    Proof : We know, adjacent angles of a parallelogram are supplementary.


    $$∠A + ∠D = 180°\\ or\space \frac{1}{2}\angle A+\frac{1}{2}\angle D=\frac{1}{2}×180° [Dividing both sides by 2]\\ ⇒ ∠QAD + ∠ADQ = 90° …(i)\\ Now, in \bigtriangleup ADQ\\ ∠ADQ + ∠QAD + ∠AQD = 180° [Angle sum property of a \bigtriangleup]\\ ⇒ 90° + ∠AQD = 180° [Using (i)]\\ ⇒ ∠AQD = 180° – 90° = 90°\\ So,∠PQR = ∠AQD = 90° [Vertically opposite angles]\\ Similarly, ∠PSR = ∠QPS = ∠QRS = 90°\\ \text{So, in quadrilateral PQRS,}\\ ∠Q = ∠S = 90° and ∠P = ∠R = 90°\\ \text{i.e., opposite angles are equal.}\\ \text{Hence, PQRS is a parallelogram.}\\ Also, ∠P = ∠Q = ∠R = ∠S = 90°\\ \text{So, PQRS is a parallelogram with each angle equal to 90°.}\\ \text{PQRS is a rectangle.}$$

    Q. In given DABC, lines are drawn through A, B and C parallel respectively to the sides BC, CA and AB forming DPQR. Show that BC = 1/2QR.


    $$\text{Proof : In quadrilateral BCQA, we have}\\ \text{BC || AQ and QC || AB}\\ \text{BCQA is a parallelogram.}\\ BC = QA \text{[Opposite sides of a ||gm] …(i)}\\ \text{Similarly, in quadrilateral ARBC, we have}\\ \text{To prove : P is the mid-point of CD i.e., DP = CP.}\\ \text{Proof : We have, ∠DAB = 60° and AP is the bisectors of ∠A}\\ ∠DAP =\frac{1}{2}\angle DAB=\frac{1}{2}×60°= 30°\\ Now, ∠DAB + ∠ABC = 180° \text{[Co-interior angles]}\\ ⇒ 60° + ∠ABC = 180°\\ ⇒ ∠ABC = 180° – 60° = 120°\\ ∠ADC = ∠ABC = 120°\\ In \bigtriangleup ADP,\\ ∠ADP + ∠PAD + ∠APD = 180° text{[Angle sum property of a D]}\\ ⇒ 120° + 30° + ∠APD = 180°\\ ⇒ ∠APD = 180° – 150° = 30°\\ ⇒ ∠DAP = ∠APD = 30°\\ DP = AD …(i) text{(opposite sides of equal angles)}\\ Similarly , ∠PBC =\frac{1}{2}\angle ABC=\frac{1}{2}×120°= 60°\\ ∠PCB = ∠DAB = 60°\\ ∠BPC = 60° [Angle sum property of ∠BCP]\\ ∠PBC = ∠PCB = ∠BPC = 60°\\ \bigtriangleup PBC text{is an equilateral triangle.}\\ CP = BC …(ii)\\ AD = BC [Opposite sides of ||gm ABCD]\\ DP = CP [Using (i) and (ii)]\\ ⇒ P is the mid-point of side CD.$$

    Q. In the given figure, ABCD is a trapezium in which
    AB || CD and AD = BC. Show that :
    (i) ∠A = ∠B
    (ii) ∠C = ∠D
    (iii) diagonal AC = diagonal BD
    Ans. Given : ABCD is a trapezium in which AB || DC and
    AD = BC.
    To prove : (i) ∠A = ∠B
    (ii) ∠C = ∠D
    (iii) Diagonal AC = BD
    Construction : Draw CE || DA and produce AB which meet CE at the point E.
    trapezium 3

    $$\text{Proof : (i) We have, AE || DC and AD || EC}\\ \text{ADCE is a parallelogram.}\\ ∠D = ∠E \space[Opposite angles of ||gm] …(A)\\ \text{ADCE is a parallelogram.}\\ and, AD = CE \space \text{[Opposite sides of ||gm]}\\ \text{But AD = BC [Given]}\\ BC = CE\\ ∠BEC = ∠CBE …(B)\space\text{[Opposite angles of equal sides]}\\ AD || CE\\ ∠A + ∠E = 180° [Co-interior angles]\\ ⇒ ∠A + ∠CBE = 180° [From (B)] …(C)\\ Also, ∠ABC + ∠CBE = 180° [Linear pair] …(D)\\ Subtracting equation (D) from equation (C), we get\\ ⇒ ∠A – ∠ABC = 0\\ ⇒ ∠A = ∠ABC = ∠B\\ (ii) Now, ∠D = ∠E\\ \text{[Opposite angle of ||gm ADCE]}\\ ⇒ ∠D = ∠CBE \text{[Using (B)]}\\ ⇒ ∠D = ∠BCD \text{[… ∠CBE = ∠BCD (Alternate angles)]}\\ ⇒ ∠D = ∠C\\ (iii) \space In \bigtriangleup ADB and \bigtriangleup ABC\\ ∠A = ∠B [Proved in (i)]\\ AB = AB [common side]\\ AD = BC [Given]\\ DADB \cong\bigtriangleup BCA [SAS congruency rule]\\ BD = AC (CPCT)$$

    Q. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angle, then it is a square.

    Ans. Consider quadrilateral ABCD in which AC = BD, AC ⊥ BD and diagonal AC and BD bisect each other.
    OA = OC and OB = OD
    To prove : ABCD is a square.


    Proof : In ΔAOB and ΔCOD,
    AO = CO [Given]
    BO = DO [Given]
    ∠AOB = ∠COD
    [Vertically opposite angles]
    ΔAOB  ≅ ΔCOD [SAS congruency rule]
    AB = CD [CPCT] ...(i)
    and ∠ABO = ∠ODC [C.P.C.T.]
    Angles ABO and ODC form a pair of alternate angles between the lines AB and DC.
    AB || DC ...(ii)
    From (i) and (ii),
    AB || DC and AB = DC
    i.e., a pair of opposite sides is parallel and equal.
    ABCD is a parallelogram. ...
    (iii) Now, in DAOB and DAOD, AO = AO [Common side]
    ∠AOB = ∠AOD [Each 90°]
    BO = DO [Given]
    ΔAOB  ≅ ΔAOD [SAS congruency rule]
    AB = AD [C.P.C.T.]
    i.e., in parallelogram ABCD, adjacent sides are equal.
    ABCD is a rhombus.
    Now, in DABD and DBAC,
    AB = AB [Common side]
    AD = BC [Proved above]
    BD = AC [Given]
    ΔABD  ≅ ΔBAC [SSS congruency rule]
    ∠BAD = ∠ABC [C.P.C.T.]
    But ∠ABC + ∠BAD = 180° [Co-interior angles]
    ⇒ 2∠ABC = 180°
    ⇒ ∠ABC = 90°
    i.e., An angle of rhombus ABCD is a right angle.
    ABCD is a square.

    Q. In the given figure, ABCD is a parallelogram, E and F are the mid-points of the sides AB and CD, respectively. Prove that the line segments AF and CE trisect the diagonal BD.

    parallelogram 1

    Q. Show that the quadrilateral formed by joining the mid-points of the consecutive sides of a rectangle is a rhombus.

    Q. In DABC, AD is the median through point A and E is the mid-point of AD. BE produced meets AC is F. Show that


    Q. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD. Show that ∠A > ∠C and ∠B > ∠D.

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