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  1. Asked: In:

    Diagonal AC of a parallelogram ABCD bisects ∠A see Figure. Show that (i) it bisects ∠C also, (ii) ABCD is a rhombus.

    amitpaswan09

    (i) ∠DAC = ∠BAC ...(1) [∵ Given] ∠DAC = ∠BCA ...(2) [∵ Alternate angle] ∠BAC = ∠ACD ...(3) [∵Alternate angles] From the equations (1), (2) and (3), we have ∠ACD = ∠BCA ...(4) Hence, Diagonal AC bisects angle C also. (ii) From the equation (2) and (4), we have ∠ACD = ∠DAC In ΔADC, ∠ACD = ∠DAC [∵ ProvRead more

    (i) ∠DAC = ∠BAC …(1) [∵ Given]
    ∠DAC = ∠BCA …(2) [∵ Alternate angle]
    ∠BAC = ∠ACD …(3) [∵Alternate angles]
    From the equations (1), (2) and (3), we have
    ∠ACD = ∠BCA …(4)
    Hence, Diagonal AC bisects angle C also.
    (ii) From the equation (2) and (4), we have
    ∠ACD = ∠DAC
    In ΔADC,
    ∠ACD = ∠DAC [∵ Prove above]
    AD = DC [∵ In a triangle, the sides opposite to equal angle are equal]
    A parallelogram whose adjacent sides are equal, is a rhombus. Hence, ABCD is a rhombus.

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  2. Asked: In:

    In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ see Figure. Show that: (i) ∆ APD ≅ ∆ CQB (ii) AP = CQ (iii) ∆ AQB ≅∆ CPD (iv) AQ = CP (v) APCQ is a parallelogram

    amitpaswan09

    (i) In ΔAPD and ΔCQB, DP = BQ [∵ Given] ∠ADP = ∠CBQ [∵ Alternate angle] AD = BC [∵ Opposite sides of a parallelogram] Hence, ΔAPD ≅ ΔCQB [∵ SAS Congruency rule] (ii) ΔAPD ≅ CQB [∵ Prove above] AP = CQ ...(1) [∵ CPCT] (iii) In ΔAQB and ΔCPD, QB = DB [∵ Given] ∠ABQ = ∠CDP [∵ Alternate angle] AB = CD [Read more

    (i) In ΔAPD and ΔCQB,
    DP = BQ [∵ Given]
    ∠ADP = ∠CBQ [∵ Alternate angle]
    AD = BC [∵ Opposite sides of a parallelogram]
    Hence, ΔAPD ≅ ΔCQB [∵ SAS Congruency rule]

    (ii) ΔAPD ≅ CQB [∵ Prove above]
    AP = CQ …(1) [∵ CPCT]

    (iii) In ΔAQB and ΔCPD,
    QB = DB [∵ Given]
    ∠ABQ = ∠CDP [∵ Alternate angle]
    AB = CD [∵ Opposite sides of a parallelogram]
    Hence, ΔAQB ≅ ΔCPD [ SAS Congruency rule]

    (iv) ΔAQB ≅ ΔCPD [∵ Prove above]
    AQ = CP …(2) [∵ CPCT]

    (v) In APCQ,
    AP = CQ [∵ From (1)]
    AQ = CP [∵ From (2)]
    The opposite sides of quadrilateral APCQ are equal.
    Hence, APCQ is a parallelogram.

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    • 7
  3. Asked: In:

    ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD see Figure. Show that (i) ∆ APB ≅ ∆ CQD (ii) AP = CQ

    amitpaswan09

    (i) In ΔAPB and ΔCQD, ∠APB = ∠CQD [∵ Each 90°] ∠ABP = ∠CDQ [∵ Alternate angles] AB = CD [∵ Opposite sides of a parallelogram] Hence, ΔAPB ≅ ΔCQD [∵ SAS Congruency rule] (ii) ΔAPB ≅ ΔCQD [∵ Prove above] AP = CQ [∵ CPCT]

    (i) In ΔAPB and ΔCQD,
    ∠APB = ∠CQD [∵ Each 90°]
    ∠ABP = ∠CDQ [∵ Alternate angles]
    AB = CD [∵ Opposite sides of a parallelogram]
    Hence, ΔAPB ≅ ΔCQD [∵ SAS Congruency rule]

    (ii) ΔAPB ≅ ΔCQD [∵ Prove above]
    AP = CQ [∵ CPCT]

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    • 4
  4. Asked: In:

    In ∆ ABC and ∆ DEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that (i) quadrilateral ABED is a parallelogram (ii) quadrilateral BEFC is a parallelogram (iii) AD || CF and AD = CF (iv) quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ∆ ABC ≅ ∆ DEF.

    amitpaswan09

    (i) In ABED, AB = DE [∵ Given] AB ∥ DE [∵ Given] Hence, ABED is a parallelogram. (ii) In BEFC, BC = EF [∵ Given] BC ∥ EF [∵ Given] Hence, BEFC is a parallelogram. (iii) In ABED, AD = BE ...(1) [∵ ABED is a parallelogram] AD ∥ BE ...(2) [∵ ABED is a parallelogram] In BEFC, BE = CF ...(3) [∵ ABED is aRead more

    (i) In ABED, AB = DE [∵ Given]
    AB ∥ DE [∵ Given]
    Hence, ABED is a parallelogram.

    (ii) In BEFC, BC = EF [∵ Given]
    BC ∥ EF [∵ Given]
    Hence, BEFC is a parallelogram.

    (iii) In ABED,
    AD = BE …(1) [∵ ABED is a parallelogram]
    AD ∥ BE …(2) [∵ ABED is a parallelogram]
    In BEFC,
    BE = CF …(3) [∵ ABED is a parallelogram]
    BE ∥ CF …(4) [∵ ABED is a parallelogram]
    From (2) and (4), we have
    AD ∥ CF …(5)
    From (1) and (3), we have
    AD = CF …(6)
    (iv) In ACFD,
    AD = CF [∵ From (6)]
    AD ∥ CF [∵ From (5)]
    Hence, ACFD is a parallelogram
    (v) In ACFD,
    AC = DF . [∵ ACFD is a parallelogram]

    (vi) In ΔABC and ΔDEF
    AB = DE [∵ Given]
    AC = DF [∵ Proved above]
    BC = EF [∵ Given]
    Hence, ΔABC ≅ ΔDEF [∵ SSS Congruency rule]

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    • 5
  6. Asked: In:

    P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC).

    amitpaswan09

    Triangle ABP and parallelogram ABCD lie on the same base AB and between the same parallels, AB ∥ CD Hence, ar (APB) = (1/2) ar (ABCD) ....(1) [∵ If a parallelogram and a triangle are on the same base and between the same parallels, than area of the triangle is half the area of the parallelogram.] SiRead more

    Triangle ABP and parallelogram ABCD lie on the same base AB and between the same parallels, AB ∥ CD
    Hence, ar (APB) = (1/2) ar (ABCD) ….(1)
    [∵ If a parallelogram and a triangle are on the same base and between the same parallels, than area of the triangle is half the area of the parallelogram.]
    Similarly,
    Triangle BQC and Parallelogram ABCD lie on the same base AB and between same parallels, AD ∥ BC.
    Hence, ar(BQC = (1/2)ar(ABCD) …(2)
    From the equation (1) and (2), ar (APB) = ar(BQC).

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