Home/ sanjeev rao/Best Answers
  1. Asked: In:

    ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Figure). Show that these altitudes are equal.

    sanjeev rao

    In ΔABE and ΔACF, ∠AEB = ∠AFC [∵ Each 90°] ∠A = ∠A [∵ Common] AB = AC [∵ Given] Hance, ΔABE ≅ ΔACF [∵ SAS Congruency Rule] BE = CF [∵ CPCT]

    In ΔABE and ΔACF,
    ∠AEB = ∠AFC [∵ Each 90°]
    ∠A = ∠A [∵ Common]
    AB = AC [∵ Given]
    Hance, ΔABE ≅ ΔACF [∵ SAS Congruency Rule]
    BE = CF [∵ CPCT]

    See less
    • 4
  2. Asked: In:

    In Δ ABC, AD is the perpendicular bisector of BC (see Figure). Show that Δ ABC is an isosceles triangle in which AB = AC.

    sanjeev rao

    In ΔABD and ΔACD, BD = DC [∵ AD bisects BC] ∠ADB = ∠ADC [∵ Each 90°] AD = AD [∵ Common] Hance, ΔABD ≅ ΔACD [∵ SAS Congruency Rule] AB = AC [∵ CPCT] Hence, ABC is an isosceles triangle.

    In ΔABD and ΔACD,
    BD = DC [∵ AD bisects BC]
    ∠ADB = ∠ADC [∵ Each 90°]
    AD = AD [∵ Common]
    Hance, ΔABD ≅ ΔACD [∵ SAS Congruency Rule]
    AB = AC [∵ CPCT]
    Hence, ABC is an isosceles triangle.

    See less
    • 5
  3. Asked: In:

    ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Figure). Show that (i) Δ ABE ≅ Δ ACF (ii) AB = AC, i.e., ABC is an isosceles triangle.

    sanjeev rao

    (i) In ΔABE and ΔACF, ∠AEB = ∠AFC [∵ Each 90] ∠A = ∠A [∵ Common] BE = CF [∵ Given] Hance, ΔABE ≅ ΔACF [∵ SAS Congruency Rule] (ii) In ΔABE ΔACF [∵ Proved above] AB = AC [∵ CPCT] Hence, ΔABC is an isosceles triangle.

    (i) In ΔABE and ΔACF,
    ∠AEB = ∠AFC [∵ Each 90]
    ∠A = ∠A [∵ Common]
    BE = CF [∵ Given]
    Hance, ΔABE ≅ ΔACF [∵ SAS Congruency Rule]

    (ii) In ΔABE ΔACF [∵ Proved above]
    AB = AC [∵ CPCT]
    Hence, ΔABC is an isosceles triangle.

    See less
    • 3
  4. Asked: In:

    ABC and DBC are two isosceles triangles on the same base BC (see Figure). Show that ∠ ABD = ∠ ACD.

    sanjeev rao

    In ΔABC, AB = AC [∵ Given] ∠ABC = ∠ACB ...(1) [∵ Angles opposite to equal sides are equal] In ΔDBC, DB = DC [∵ Given] ∠DBC = ∠DCB ...(2) [∵ Angles opposite to equal sides are equal] Adding equation (1) and (2), we get ∠ABC + ∠DBC = ∠ACB + ∠DCB ⇒ ∠ABD = ∠ACD.

    In ΔABC,
    AB = AC [∵ Given]
    ∠ABC = ∠ACB …(1) [∵ Angles opposite to equal sides are equal]
    In ΔDBC,
    DB = DC [∵ Given]
    ∠DBC = ∠DCB …(2) [∵ Angles opposite to equal sides are equal]
    Adding equation (1) and (2), we get
    ∠ABC + ∠DBC = ∠ACB + ∠DCB
    ⇒ ∠ABD = ∠ACD.

    See less
    • 3
  6. Asked: In:

    ΔABC is an isosceles triangle in which AB = AC.Side BA is produced to D such that AD = AB (see Figure). Show that ∠ BCD is a right angle.

    sanjeev rao

    In ΔACD, AB = AC [∵ Given] ∠ACD = ∠D ...(1) [∵ Angles opposite to equal sides are equal] In ΔABC, AB = AC [∵ Given] ∠B = ∠ACB ...(2) [∵ Angles opposite to equal sides are equal] In ΔDBC, ∠D + ∠B + ∠BCD = 180° ⇒ ∠ACD + ∠ACB + BCD = 180° [∵ From the equation (1) and (2)] ⇒ ∠BCD + ∠BCD = 180° [∵ ∠ACD +Read more

    In ΔACD,
    AB = AC [∵ Given]
    ∠ACD = ∠D …(1) [∵ Angles opposite to equal sides are equal]
    In ΔABC,
    AB = AC [∵ Given]
    ∠B = ∠ACB …(2) [∵ Angles opposite to equal sides are equal]
    In ΔDBC,
    ∠D + ∠B + ∠BCD = 180°
    ⇒ ∠ACD + ∠ACB + BCD = 180° [∵ From the equation (1) and (2)]
    ⇒ ∠BCD + ∠BCD = 180° [∵ ∠ACD + ∠ACB = ∠BCD]
    ⇒ 2∠BCD = 180°
    ⇒ ∠BCD = (180°/2) = 90°
    Hence, ∠BCD is a is a right angle.

    See less
    • 2
  7. 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°.

    See less
    • 2
  8. 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]

    See less
    • 2
  9. 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.

    See less
    • 1
  11. 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]

    See less
    • 2
  12. 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]

    See less
    • 4