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  1. Asked: In:

    Show that the angles of an equilateral triangle are 60° each.

    sanjeev rao

    In ΔABC, AB = AC [∵ Given] ∠C = ∠B ...(1) [∵ Angles opposite to equal sides are equal] Similarly, In ΔABC, AB = BC [∵ Given] ∠C = ∠A ...(2) [∵ Angles opposite to equal sides are equal] From the equation (1) and (2), we have ∠A = ∠B = ∠C ...(3) In ΔABC, ∠A + ∠B + C = 180° ⇒ ∠A +∠A+ ∠A= 180° [∵ From tRead more

    In ΔABC,
    AB = AC [∵ Given]
    ∠C = ∠B …(1) [∵ Angles opposite to equal sides are equal]
    Similarly,
    In ΔABC,
    AB = BC [∵ Given]
    ∠C = ∠A …(2) [∵ Angles opposite to equal sides are equal]
    From the equation (1) and (2), we have

    ∠A = ∠B = ∠C …(3)
    In ΔABC,
    ∠A + ∠B + C = 180°
    ⇒ ∠A +∠A+ ∠A= 180° [∵ From the equation (3)]
    ⇒ 3∠A = 180°
    ⇒ ∠A = (180°/3) = 60°
    Hence, ∠A = ∠B = ∠C = 60°.

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  2. Asked: In:

    Δ ABC and Δ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Figure). If AD is extended to intersect BC at P, show that (i) Δ ABD ≅ Δ ACD (ii) Δ ABP ≅ Δ ACP (iii) AP bisects ∠ A as well as ∠ D. (iv) AP is the perpendicular bisector of BC.

    sanjeev rao

    In ΔABD and ΔACD, AB = AC [∵ Given] BD = CD [∵ Given] AD = AD [∵ Common] Hence, ΔABD ≅ ΔACD [∵ SSS Congruency Rule] (ii) In ΔABD ≅ ΔACD, [∵ Proved above] ∠BAD = ∠CAD [∵ CPCT] In ΔABP and ΔACP, AB = AC [∵ Given] ∠BAP = ∠CAP [∵ Proved above] AP = AP [∵ Common] Hence, ΔABP ≅ ΔACP [∵ SAS Congruency RuleRead more

    In ΔABD and ΔACD,
    AB = AC [∵ Given]
    BD = CD [∵ Given]
    AD = AD [∵ Common]
    Hence, ΔABD ≅ ΔACD [∵ SSS Congruency Rule]

    (ii) In ΔABD ≅ ΔACD, [∵ Proved above]
    ∠BAD = ∠CAD [∵ CPCT]
    In ΔABP and ΔACP,
    AB = AC [∵ Given]
    ∠BAP = ∠CAP [∵ Proved above]
    AP = AP [∵ Common]
    Hence, ΔABP ≅ ΔACP [∵ SAS Congruency Rule]

    (III) In ΔABD ≅ ΔACD [∵ Proved above]
    ∠BAD = ∠CAD [∵ CPCT]
    ∠BDA = ∠CDA [∵ CPCT]
    Hence, AP bisects both the angles A and D.

    (iv) In ΔABP ≅ ΔACP [∵ Proved above]
    BP = CP [∵ CPCT]
    ∠BPA = ∠CPA [∵ CPCT]
    ∠BPA + ∠CPA = 180° [∵ Linear pair]
    ⇒∠CPA + ∠CPA = 180° [∵∠BPA = ∠CPA]
    ⇒ 2∠CPA = 180° ⇒∠CPA = (180°/2 = 90°
    ⇒ AP is perpendicular to BC. ⇒ AP is Perpendicular bisector of BC. [∵ BP = CP]

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  3. Asked: In:

    AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that (i) AD bisects BC (ii) AD bisects AngleA.

    sanjeev rao

    (i) In ΔABD and ΔACD, ∠ADB = ∠ADC [∵ Each 90°] AB = AC [∵ Given] AD = AD [∵ Common] Hence, ΔABD ≅ ΔACD [∵ RHS Congruency Rule] BD = DC [∵ CPCT] Hence, AD bisects BC. (ii) ∠BAD = ∠CAD [∵ CPCT] Hence, AD bisects angle A.

    (i) In ΔABD and ΔACD,
    ∠ADB = ∠ADC [∵ Each 90°]
    AB = AC [∵ Given]
    AD = AD [∵ Common]
    Hence, ΔABD ≅ ΔACD [∵ RHS Congruency Rule]
    BD = DC [∵ CPCT]
    Hence, AD bisects BC.

    (ii) ∠BAD = ∠CAD [∵ CPCT]
    Hence, AD bisects angle A.

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  4. Asked: In:

    Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of Δ PQR (see Figure). Show that: (i) Δ ABM ≅ Δ PQN (ii) Δ ABC ≅ Δ PQR

    sanjeev rao

    (i) Given that : BC = QR ⇒ 1/2 BC = 1/2QR ⇒ BM = QN [∵ AM and PM are medians] In ΔABM and ΔPQN, AB = PQ [∵ Given] AM = PN [∵ Given] BM = QN [∵ Proved above] Hence, ΔABM ≅ ΔPQN [∵ SSS Congruency Rule] (ii) In ΔABM ≅ ΔPQN, [∵ Proved above] ∠B = ∠Q [∵ CPCT] In ΔABC and ΔPQR, AB = PQ [∵ Given] ∠B = ∠Q [Read more

    (i) Given that : BC = QR
    ⇒ 1/2 BC = 1/2QR ⇒ BM = QN [∵ AM and PM are medians]
    In ΔABM and ΔPQN,
    AB = PQ [∵ Given]
    AM = PN [∵ Given]
    BM = QN [∵ Proved above]
    Hence, ΔABM ≅ ΔPQN [∵ SSS Congruency Rule]

    (ii) In ΔABM ≅ ΔPQN, [∵ Proved above]
    ∠B = ∠Q [∵ CPCT]
    In ΔABC and ΔPQR,
    AB = PQ [∵ Given]
    ∠B = ∠Q [∵ Proved above]
    BC = QR [∵ Given]
    Hence, ΔABC ≅ ΔPQR [∵ SAS Congruency Rule]

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  6. 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.

    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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