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  1. Asked: In:

    ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see Figure). Prove that (i) ΔABD ≅ ΔBAC (ii) BD = AC (iii) ∠ABD = ∠BAC.

    Best Answer
    sanjeev rao

    In ΔABD and ΔBAC, AD = BC [∵ Given] ∠DAB = ∠CAB [∵ Given] AB = AB [∵ Common] Hance, ΔABD ≅ ΔBAC [∵ SAS Congruency Rule] (ii) BC = BD [∵ Corresponding parts of congruent triangles are equal] (iii) ∠ABD = ∠BAC [∵ Corresponding parts of congruent triangles are equal]

    In ΔABD and ΔBAC,
    AD = BC [∵ Given]
    ∠DAB = ∠CAB [∵ Given]
    AB = AB [∵ Common]
    Hance, ΔABD ≅ ΔBAC [∵ SAS Congruency Rule]
    (ii) BC = BD [∵ Corresponding parts of congruent triangles are equal]
    (iii) ∠ABD = ∠BAC [∵ Corresponding parts of congruent triangles are equal]

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  2. Asked: In:

    AD and BC are equal perpendiculars to a line segment AB (see Figure). Show that CD bisects AB.

    Best Answer
    sanjeev rao

    In ΔOCB and ΔODA, ∠BOC = ∠AOD [∵ Vertically Opposite Angles] ∠CBO = ∠DAO [∵ Each 90°] BC = AD [ Given] Hance, ΔOCB ≅ ΔODA [∵ AAS Congruency Rule] BO = AO [∵ Corresponding parts of congruent triangles are equal] Hence. CD bisects AB.

    In ΔOCB and ΔODA,
    ∠BOC = ∠AOD [∵ Vertically Opposite Angles]
    ∠CBO = ∠DAO [∵ Each 90°]
    BC = AD [ Given]
    Hance, ΔOCB ≅ ΔODA [∵ AAS Congruency Rule]
    BO = AO [∵ Corresponding parts of congruent triangles are equal]
    Hence. CD bisects AB.

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  3. Asked: In:

    l and m are two parallel lines intersected by another pair of parallel lines p and q (see Figure). Show that ΔABC ≅ ΔCDA.

    Best Answer
    sanjeev rao

    In ΔABC and ΔCDA, ∠BAC = ∠ACD [∵ Alternate Angles] AC = AC [∵ Common] ∠BCA = ∠DAC [∵ Alternate Angles] Hance, ΔABC ≅ ΔCDA [∵ ASA Congruency Rule]

    In ΔABC and ΔCDA,
    ∠BAC = ∠ACD [∵ Alternate Angles]
    AC = AC [∵ Common]
    ∠BCA = ∠DAC [∵ Alternate Angles]
    Hance, ΔABC ≅ ΔCDA [∵ ASA Congruency Rule]

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  4. Asked: In:

    line l is the bisector of an angle ∠ A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠ A (see Figure). Show that: (i) Δ APB ≅ Δ AQB (ii) BP = BQ or B is equidistant from the arms of ∠ A.

    Best Answer
    sanjeev rao

    (i) In ΔAPB and ΔAQB, ∠APB = ∠AQB [∵ Each 90°] ∠PAB = ∠QAB [∵ Line l bisects angle A] AB = AB [∵ Common] Hance, ΔAPB ≅ ΔAQB [∵ ASA Congruency Rule] (ii) BP = BQ [∵ Corresponding parts of congruent Triangles are equal ]

    (i) In ΔAPB and ΔAQB,
    ∠APB = ∠AQB [∵ Each 90°]
    ∠PAB = ∠QAB [∵ Line l bisects angle A]
    AB = AB [∵ Common]
    Hance, ΔAPB ≅ ΔAQB [∵ ASA Congruency Rule]

    (ii) BP = BQ [∵ Corresponding parts of congruent Triangles are equal ]

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  6. Asked: In:

    In Figure, AC = AE, AB = AD and ∠ BAD = ∠ EAC. Show that BC = DE.

    Best Answer
    sanjeev rao

    ∠BAD = ∠EAC [∵ Given] Adding ∠DAC both sides, we have ∠BAD + ∠DAC = ∠EAC + ∠DAC ⇒ ∠BAC = ∠EAD In ΔBAC and ΔDAE, AB = AD [∵ Given] ∠BAC = ∠EAD [∵ Proved above] AC = AE [∵ Given] Hance, ΔBAC ≅ ΔDAE [∵ SAS Congruency Rule] BC = DE [∵ Corresponding parts of congruent Triangles are equal ]

    ∠BAD = ∠EAC [∵ Given]
    Adding ∠DAC both sides, we have
    ∠BAD + ∠DAC = ∠EAC + ∠DAC
    ⇒ ∠BAC = ∠EAD
    In ΔBAC and ΔDAE,
    AB = AD [∵ Given]
    ∠BAC = ∠EAD [∵ Proved above]
    AC = AE [∵ Given]
    Hance, ΔBAC ≅ ΔDAE [∵ SAS Congruency Rule]
    BC = DE [∵ Corresponding parts of congruent Triangles are equal ]

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    • 2